2 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip

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50 questions · 1 auto-graded MCQ + 49 self-marked written.

Question 12 Marks

At absolute zero, an exothermic reaction is always spontaneous but at temperatures above absolute zero, we have to consider both enthalpy and entropy before we can predict spontaneity. Why?

Answer

As ∆G = ∆H –T∆S
The process is spontaneous when ∆G is –ve. Since T = 0 so T∆S = 0, i.e ∆G = ∆H. For an exhothermic process, ∆H is –ve, therefore at absolute zero ∆G will always be –ve and hence a spontaneous process.
At temp. above absolute zero, ∆S is not zero. It may be +ve or –ve. Hence we have to consider both the ∆H and T∆S for deciding ∆G and the spontaneity.

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Question 22 Marks

Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of $K_H$ (Henry's constant) and why?
In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?

Answer

Gas B, Higher the value of $K_H$ lower is the solubility of gas/$\text{P} = \text{K}_{H}\text{x}$.
Negative deviation from Raoult’s law.

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Question 32 Marks

Why are aquatic species more comfortable in cold water than in warm water?
What happens when we place the blood cell in saline water solution (hypertonic solution)? Give reason.

Answer

As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lakes/as cold water contains more oxygen dissolved.
They will shrink, due to osmosis.

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Question 42 Marks

State the following:

Raoult's law in its general form in reference to solutions.
Henry's law about partial pressure of a gas in a mixture.

Answer

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution.

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Question 52 Marks

Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?

Answer

The flow of solvent from solution of low concentration to higher concentration through semipermeable membrane is called osmosis. The hydrostatic pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the Osmotic Pressure.
Advantage: Unlike other colligative properties, osmotic pressure is used to determine the Molar mass of macromolecules/polymers like protein.

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Question 62 Marks

Define the term ‘osmotic pressure’. Describe how the molecular mass of a substance can be determined on the basis of osmotic pressure measurement.

Answer

The extra pressure applied on the solution side that just stops the flow of solvent to solution through semi-permeable membrane is called osmotic pressure of the solution.
Here r is the osmotic pressure and R is the gas constant.
$\pi = (n2/V) RT$
$\pi\text{ V}=\frac{\text{w}_{2}\text{ RT}}{\text{M}_{2}}$
or $\text{ M}_{2}=\frac{\text{w}_{2}\text{ RT}}{\pi\text{ V}}$
Thus knowing the quantities $w_2, T,\eth$ and V we can calculate the molar mass of the solute.

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Question 72 Marks

State Raoult’s law for solutions of volatile liquid components. Taking a suitable example, explain the meaning of positive deviation from Raoult’s law.

Answer

Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proprtional to its mole fraction.
When the solute-solvent interaction is weaker than those between the solute-solute and solvent-solvent molecules then solution shows positive deviation from Raoults law hence the partial pressure of each component is greater. ex. mixture of ethanol and acetone or carbondisulphide and acetone behave in this manner.

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Question 82 Marks

What is meant by elevation in boiling point? Why is it a colligative property?

Answer

The increase in boiling point of the solvent in a solution when a non-volatile solute is added.
Because it depends upon molality/the number of solute particles rather than their nature/$\triangle\text{T}_{\text{b}}\propto\text{m}$

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Question 92 Marks

State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid?

Answer

Henry’s law states that the mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
Applications: solubility of $\text{CO}_2$ gas in soft drinks/solubility of air diluted with helium in blood used by sea divers or any other.
Solubility of gas in liquid decreases with increase in temperature.

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Question 102 Marks

State Raoult’s law for a solution containing non-volatile solute. What type of deviation from Raoult’s law is shown by a solution of chloroform and acetone and why?

Answer

The relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute. / The vapour pressure of a solution of a non- volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.
Negative deviation due to formation of Hydrogen bond between chloroform and acetone.

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Question 112 Marks

Define the following terms:

Abnormal molar mass.
van’t Hoff factor (i).

Answer

If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass
Extent of dissociation or association or ratio of the observed colligative property to calculated colligative property

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Question 122 Marks

State Raoult's law for the solution containing volatile components. What is the similarity between Raoult's law and Henry's law?

Answer

For the solution containing volatile components, the partial vapour pressure of each component is directly proportional to its mole fraction. In both cases, p ∝ x/ Henry’s Law is a special case of Raoults Law.

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Question 132 Marks

Define the following terms :

Ideal solution.
Molarity $(M).$

Answer

solution:

The solution which obey Roult's law over entire range of concentration are known as ideal solution.
For ideal solution: Enthalpy of mixing of the pure components to form the solution, $\triangle _{mix}\ H = 0$ and volume of mixing, $\triangle _{mix}\ V = 0$

Molarity:

It is defined as the number of moles of solute present in $1000$ mL of the solution. Molarity is represented by $M.$
$\text{Molarity (M)}=\frac{\text{Number of moles of solute}}{\text{ volume of solution in mL}}\times1000$
or
$\text{M} = \frac{\text{Weight of solute}}{\text{molar mass of solute}\times \text{Volume of solution in mL}}\times1000$

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Question 142 Marks

Define the following terms:

Colligative properties.
Molality (m).

Answer

Properties that are independent of nature of solute and depend on number of moles of solute only.
Number of moles of solute dissolved per kg of the solvent.

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Question 152 Marks

What is meant by positive deviations from Raoult's law? Give an example. What is the sign of $\triangle _{mix}H$ for positive deviation?

Answer

When vapour pressure of solution is higher than that predicted by Raoult's law/the intermolecular attractive forces between the solute-solvent/(A-B) molecules are weaker than those between the solute-solute and solvent-solvent molecules/A-A or B-B molecules.
Eg. ethanol-acetone/ethanol-cyclohexane$/CS_2$-acetone.

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Question 162 Marks

Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult's law? Give an example.

Answer

Azeotropes are binary mixtures having the same composition in the liquid and vapour phase and boil at a constant temperature.
Minimum boiling azeotrope.

eg - ethanol+water.

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Question 172 Marks

Define an ideal solution and write one of its characteristics.

Answer

Solutions which obey Raoult’s law over the entire range of concentration:
$\mathrm{A}-\mathrm{A}$ or $\mathrm{B}-\mathrm{B} \sim \mathrm{A}-\mathrm{B}$ interactions.
$\Delta \mathrm{H}_{\text {mix }}=0$
$\Delta \mathrm{~V}_{\text {mix }}=0$

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Question 182 Marks

Calculate the mass of compound $($molar mass $=256 \mathrm{~g} \mathrm{~mol}^{-1} )$ to be dissolved in $75$ g of benzene to lower its freezing point by $0.48\ K\ (K_f= 5.12\ K\ kg\ mol^{-1}).$

Answer

$\Delta\text{T}_{f}=\frac{\text{K}_{f}\times\text{w}_{2}\times\text{1000}}{\text{w}_{1}\times\text{M}_{2}}$
$\text{0.48K=5.12Kkgmol}^{-1}\times\frac{\text{W}_{2}}{\text{75}\times{256}}\times1000$
$\text{W}_{2}=\frac{\text{0.48}\times\text{75}\times\text{256}}{\text{5.12}\times\text{1000}}$
$w_2= 1.8g$

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Question 192 Marks

$18g$ of glucose, $C_6H_{12}O_6($Molar Mass$=180g\ mol^{-1})$ is dissolved in $1\ kg$ of water in a sauce pan. At what temperature will this solution boil?
$(K_b$ for water $= 0.52\ K\ kg\ mol^{-1},$ boiling point of pure water $= 373.15\ K).$

Answer

$ \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{~m} . $
$ \mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}{ }^0=0.52\ \mathrm{K\ kg}\ \mathrm{~mol}^{-1} \times \frac{18 \mathrm{~g}}{180 \mathrm{gmol}^{-1}} \times \frac{1}{1 \mathrm{~kg}} $
$ \mathrm{~T}_{\mathrm{b}}-373.15 \mathrm{~K}=0.052 \mathrm{~K} .$
$ \mathrm{T}_{\mathrm{b}}=373.202 \mathrm{~K} .$

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Question 202 Marks

A $1.00$ molal aqueous solution of trichloroacetic acid $\left(\mathrm{CCl}_3 \mathrm{COOH}\right)$ is heated to its boiling point. The solution has the boiling point of $100.18{ }^{\circ} \mathrm{C}$. Determine the van't Hoff factor for trichloroacetic acid. $\left(\mathrm{K}_{\mathrm{b}}\right.$ for water $\left.=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$.

Answer

$ \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \mathrm{~K}_{\mathrm{b}} \mathrm{~m}$
$ (100.18-100)^{\circ} \mathrm{C}=\mathrm{i} \times 0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1 \mathrm{~m} $
$ 0.18 \mathrm{~K}=\mathrm{i} \times 0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1 \mathrm{~m} $
$ \mathrm{i}=0.35$

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Question 212 Marks

Define the following terms:

Mole fraction.
Isotonic solutions.
Van't Hoff factor.
Ideal solution.

Answer

Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
Two solutions having same osmotic pressure at a given temperature are called Isotonic Solutions.
Van’t Hoff factor is expressed as:

$\text{i}=\frac{\text{Normal molar mass}}{\text{abnormal molar mass}}$

The solution which obeys Raoult’s law under all conditions is known as an ideal solution.

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Question 222 Marks

Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type.

Answer

When the vapour pressure of a non-ideal solution is either higher or lower than that predicted by Raoult’s law, the solution exhibits deviations. These deviations are caused because of unequal intermolecular attractive forces between solute-solvent molecules and solute-solute or solvent-solvent molecules. Positive deviation eg: Mixture of ethanol and acetone, Carbon-disulphide and acetone. Negative deviation eg: Chloroform and acetone, nitric acid and water.

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Question 232 Marks

Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity?

Answer

Molality (m) is the number of moles of the solute per kilogram (kg) of the solvent whereas Molarity is the number of moles of solute present in one litre (or one cubic decimeter) of solution.
Molality is independent of temperature whereas Molarity is function of temperature because volume depends on temperature and the mass does not or Molarity decreases with increase in temprature.

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Question 242 Marks

State Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention two applications for the law.

Answer

Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution.
Applications:

To increase the solubility of $\text{CO}_2$ in soft drinks and soda water, the bottle is sealed under high pressure.
Scuba divers must cope with high concentrations of dissolved Nitrogen with breathing air at high pressure underwater. To avoid this air is diluted with He.
At high altitudes the partial pressure of oxygen is less than that at the ground level. Low blood oxygen causes anoxia.

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Question 252 Marks

Calculate the freezing point of a solution containing $60\ g$ of glucose $($Molar mass $=180 \mathrm{~g} \mathrm{~mol}^{-1}$ $)$ in $250\ g$ of water. $\left(K_f\right.$ of water $\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

Answer

Given,
$\mathrm{W}_2=60 \mathrm{~g} $
$ \mathrm{M}_2=180 \mathrm{~g} \mathrm{~mol}^{-1} $
$ \mathrm{~W}_1=250 \mathrm{~g} $
$ \mathrm{~K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $
$ \Delta \mathrm{~T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}} \times \mathrm{m} $
$ \mathrm{~T}_{\mathrm{f}}^0-\mathrm{T}_{\mathrm{f}}=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{w}_2 \times 1000}{\mathrm{M}_2 \times \mathrm{W}_1} $
$ 273-\mathrm{T}_{\mathrm{f}}=\frac{1.86 \times 60 \times 1000}{180 \times 250} $
$ 273.15-\mathrm{T}_{\mathrm{f}}=2.48 $
$ \mathrm{~T}_{\mathrm{f}}=273.15-2.48=270.67 \mathrm{k}$
Hence, the freezing point is $270.67\ K$ or $-2.48^{\circ} \mathrm{C}$.

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Question 262 Marks

State Henry’s law and write its two applications.

Answer

Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”:

To increase the solubility of $\text{CO}_2$ in soft drinks.
At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers.
Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood.

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Question 272 Marks

For a reaction
$2\text{H}_2\text{O}_2\frac{\text{I}^-}{\text{alkaline medium}}2\text{H}_2\text{O}+\text{O}_2$
the proposed mechanism is as given below:

$\text{H}_2\text{O}_2+\text{I}^{-}\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\text{O}+\text{IO}^{-}(\text{slow})$
$\text{H}_2\text{O}_2+\text{IO}^{-}\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\text{O}+\text{I}^{-}+\text{O}_2(\text{fast})$

Write rate law for the reaction.
Write the overall order of reaction.
Out of steps (1) and (2), which one is rate determining step?

Answer

$2\text{H}_2\text{O}_2\frac{\text{I}^-}{\text{alkaline medium}}2\text{H}_2\text{O}+\text{O}_2$In Some cases, the rate of reaction depends not only on the reactant but may also depend on the substance present as a catalyst. This is also seen in the above reaction. The rate of reaction depends on the slowest step in case of a complex reaction.

$\text{H}_2\text{O}_{2} +\text{I}^- \xrightarrow{\ \ \text{Slow}\ \ }\text{H}_{2}\text{O} + \text{IO}^-$
$\text{H}_2\text{O}_{2} +\text{IO} \xrightarrow{\ \ \text{Fast}\ \ }\text{H}_{2}\text{O} + \text{I}^-+\text{O}_2$

Rate law:

$\text{Rate}=-\frac{1}{2}\frac{\text{d}\big[\text{H}_2\text{O}_2\big]}{\text{dt}}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

The overall order of reaction:

$\text{Rate}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$
$\text{Order}=1+1=2$

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Question 282 Marks

For a $5 \%$ solution of urea $($Molar mass $=60 \mathrm{~g} / \mathrm{mol} ),$ calculate the osmotic pressure at $300 \mathrm{~K} .\left[\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Answer

$\pi=\text{CRT} ($Volume of solution $= 100mL)$
$\pi=\frac{\text{n}}{\text{v}}\text{RT}$
$\pi=\frac{5}{60}\times\frac{0.0821\times300}{0.1}$
$\pi=20.5\text{atm}.$

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Question 292 Marks

Write two differences between an ideal solution and a non-ideal solution.

Answer

Ideal	Non-ideal
Obeys Roult’s law at all range of concentration.	Does not obey.
$\Delta_\text{mix}\text{H}=0,\ \Delta_\text{mix}\text{V}=0$	$\Delta_\text{mix}\text{H}\neq0,\ \Delta_\text{mix}\text{V}\neq0$ (or any other difference)

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Question 302 Marks

Visha took two aqueous solutions - one containing 7.5g of urea (Molar mass = 60g/ mol) and the other containing 42.75g of substance Z in 100g of water, respectively. It was observed that both the solutions froze at the same temperature. Calculate the molar mass of Z.

Answer

$\triangle\text{T}_\text{f}(\text{urea})=\triangle\text{T}_\text{f}\text{(Z)}$
$\text{kf}\times\frac{\text{w urea}}{\text{Murea}}\times\frac{1000}{\text{w solvent}}$
$=\text{kf}\times\frac{\text{wz}}{\text{Mz}}\times\frac{1000}{\text{w solvent}}$
$=\frac{7.5}{60}\times\frac{1000}{100}=\frac{42.75}{\text{Mz}}\times\frac{1000}{100}$
$\text{Mz}=\frac{42.75\times60}{7.50}=342\text{g/ mol}.$

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Question 312 Marks

State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.

Answer

Rault’s law states that for a solution of volatile liquids, the partial pressure of each component is directly proportional to their mole fraction in solution.
Raoult's Law: According to Raoult's law, for a solution of two volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
$\text{P}_1\alpha\text{ x}_1$
$\Rightarrow\text{P}_1=\text{P}^{0}_1\text{x}_1$
$P_1=$ partial pressure of component $1$
$x_1=$ mole fraction of component $1$
$\text{P}^0_1$ pure vapour pressure of component $1$
Ideal solution obeys Raoult's law at all concentrations. The conditions obeyed are,

$\Delta_{\text{mix}}\text{H}=0$
$\Delta_{\text{mix}}\text{V}=0$

Two characteristics of ideal solutions:

They obey rault’s law at all ranges of concentration. I.e. Ptotal $= P_1 + P_2$
Both $\Delta\text{H}$ and $\Delta\text{V}$ are zero for such solutions.

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Question 322 Marks

For a reaction
$2\text{H}_2\text{O}_2\xrightarrow[\text{alkaline medium}]{\text{I}^-}2\text{H}_2\text{O}+\text{O}_2$
the proposed mechanism is as given below:

$\text{H}_2\text{O}_2+\text{I}^{-}\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\text{O}+\text{IO}^{-}(\text{slow})$
$\text{H}_2\text{O}_2+\text{IO}^{-}\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\text{O}+\text{I}^{-}+\text{O}_2(\text{fast})$

Write rate law for the reaction.
Write the overall order of reaction.
Out of steps (1) and (2), which one is rate determining step?

Answer

$2\text{H}_2\text{O}_2\xrightarrow[\text{alkaline medium}]{\text{I}^-}2\text{H}_2\text{O}+\text{O}_2$In some cases, the rate of reaction depends not only on the reactant but may also depend on the substance present as a catalyst. This is also seen in the above reaction. The rate of reaction depends on the slowest step in case of a complex reaction.

$\text{H}_2\text{O}_{2} +\text{I}^- \xrightarrow{\ \ \text{Slow}\ \ }\text{H}_{2}\text{O} + \text{IO}^-$
$\text{H}_2\text{O}_{2} +\text{IO}^- \xrightarrow{\ \ \text{Fast}\ \ }\text{H}_{2}\text{O} + \text{I}^-+\text{O}_2$

Rate law:

$\text{Rate}=-\frac{1}{2}\frac{\text{d}\big[\text{H}_2\text{O}_2\big]}{\text{dt}}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

The overall order of reaction:

$\text{Rate}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$
$\text{Order}=1+1=2$

Slowest step is the rate determining step. Hence, step 1 is rate determining step.

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Question 332 Marks

The vapour pressure of water is $12.3$ kPa at $300$ K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer

$1$ molar solution of solute means $1$ mole of solute in $1000g$ of the solvent.Molar mass of water $($solvent$) = 18\ g\ mol^{-1}$
$\therefore\ \text{Moles of water}=\frac{1000}{18}=55.5\text{ moles.}$
$\therefore\ \text{Mole fraction of solute}=\frac{1}{1+55.5}=0.0177$
$\text{Now},\ \frac{\text{P}^\circ-\text{P}_{\text{s}}}{\text{P}^\circ}=\text{x}_2$
$\frac{12.3-\text{P}_\text{s}}{12.3}=0.0177$
$\Rightarrow\text{P}_\text{s}=12.08\text{ KPa}$

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Question 342 Marks

Explain the following phenomena with the help of Henry’s law.
Why soda water bottle kept at room temperature fizzes on opening?

Answer

When a soda water bottle kept at room temperature is opened to air the partial pressure of $\text{CO}_2$ above the solution decreases suddenly, (as per Henry's law).This results into a decrease in solubility of carbon dioxide, hence $\text{CO}_2$ bubbles come out of the bottle with a fizz.

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Question 352 Marks

Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis.

Answer

Material used for making semipermeable membrane for carrying out reverse osmosis is- "a film of cellulose acetate placed over a suitable support."

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Question 362 Marks

At $300\ K, 36\ g$ of glucose present in a litre of its solution has an osmotic pressure of $4.98$ bar. If the osmotic pressure of the solution is $1.52 $ bars at the same temperature, what would be its concentration?

Answer

$\pi_1=\text{C}_1\text{RT}\ ....(\text{i})$$\pi_2=\text{C}_2\text{RT}\ ....(\text{ii})$
therefore on dividing the $1$ by $2$ we get
$\frac{\pi_1}{\pi_2}=\frac{\text{C}_1}{\text{C}_2}$
Let us calculate the concentration of the first solution with osmatic pressure $4. 98$bar
mass of glucose $= 36\ g$
molar mass of gulcose $= 180\ g/\ mol$
therefore number of moles of gulcose $= 36/ 180$
$= 0.2$ moles
volume of the solution $= 1\ L$
molarity = No. of moles of glucose/ vol.of solution
molarity $= 0.2/ 1\ L$
$C_1 = 0.2$ moles/ L
$\text{C}_2=\frac{\pi_2\text{C}_1}{\pi_2}$
$\text{C}_2=\frac{1.52\times0.2}{4.98}=0.61\text{ M}$

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Question 372 Marks

What will happen to freezing point of a potassium iodide aqueous solution when mercuric iodide is added to solution?

Answer

$\mathrm{Hgl}_2+2 \mathrm{KI} \rightarrow \mathrm{~K}_2 \mathrm{HgI}_4$
Mercuric iodide forms a complex with potassium iodide, therefore, the number of solute particles ( KI ) in the solution decreases resulting in the decrease in the value of $\Delta \mathrm{T}_{\mathrm{f}}$, i.e., depression in the freezing point. As a result, the freezing point $\mathrm{Hgl}_2+2 \mathrm{KI} \rightarrow \mathrm{K}_2 \mathrm{Hgl}_4$ of the solution will increase.

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Question 382 Marks

If the density of some lake water is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$ and contains 92 g of $\mathrm{Na}^{+}$ions per kg of water, calculate the molality of $\mathrm{Na}^{+}$ions in the lake.

Answer

number of moles present in $92$ g of $\mathrm{Na}^{+}$ion. $=92 / 23 \mathrm{~g} \mathrm{~mol}^{-1}=4 \mathrm{~mol}$
Molality, $\mathrm{m}=\frac{\text { No. of gm moles of solute }}{\text { wt. of solvent in } \mathrm{kg}}$
No. of gm moles of solute $=92 / 23=4$
Wt. of water (solvent) $=1 \mathrm{~kg}$
Molality $=\frac{4}{1}=4 \mathrm{~m}$.

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Question 392 Marks

State Henry’s law and mention some important applications?

Answer

The effect of pressure on the solubility of a gas in a liquid is governed by Henry's Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas Mathematically, $P = K_HX$ where $P$ is the partial pressure of the gas; and $X$ is the mole fraction of the gas in the solution and $K_H$ is Henry's Law constant.
Applications of Henry's law:

In the production of carbonated beverages $($as solubility of $C02$ increases at high pressure$).$
In the deep sea diving.
For climbers or people living at high altitudes, where low blood $0_2$ causes climbers to become weak and make them unable to think clearly.

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Question 402 Marks

What is "semi permeable membrane"?

Answer

A membrane that permits the flow of solvent molecules not the solute molecules is called semi permeable membrane. During osmosis and reverse osmosis, only solvent molecules move across the semi permeable membrane.

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Question 412 Marks

Calculate the mass of ascorbic acid $\left(\right.$ Vitamin $\left.\mathrm{C}_1 \mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_6\right)$ to be dissolved in $75$ g of acetic acid to lower its melting point by $1.5^{\circ} \mathrm{C} . \mathrm{Kf}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.

Answer

$\text{Given:} \ \Delta\text{T}_\text{f}=1.5^\circ$ Mass of $CH_3COOH,\ w_1 = 75\ g\ M_1 = 60g\ mol^{-1}M_2(C_6H_8O_6) = 176\ g\ mol^{-1}$
$K_f = 3.9K\ Kg\ mol^{-1}$
To find: $w_2 = ?$
Solution:
$\text{Applying M}_2=\frac{1000\text{K}_{\text{f}}\text{w}_2}{\text{w}_1\Delta\text{T}_\text{f}}$
$\text{or},\ \ \text{w}_2=\frac{\text{M}_2\times\text{w}_1\times\Delta\text{T}_\text{f}}{1000\times\text{K}_\text{f}}$
$\text{w}_2=\frac{176\times75\times1.5}{1000\times3.9}=5.077\text{g}$

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Question 422 Marks

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer

When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier's Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

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Question 432 Marks

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0\ g$ of polymer of molar mass $185,000$ in $450$ mL of water at $37^\circ C.$

Answer

We have given that, Volume of water, $V = 450ml = 0.450L$ Temperature, $T = (37 + 273)K = 310K$$\pi=?, \text{V}=450\text{ mL}=0.450\text{L},$
$T = 273 + 37 = 310K$ $R = 0.083\ L$ bar $mol^{-1}K^{-1},\ M_2 = 185,000,\ W_2 = 1.0\ g$
$\text{M}_2=\frac{\text{W}_2\text{RT}}{\pi\text{V}}$$185,000=\frac{1.0\times0.083\times310}{\pi\times0.450}$
$\pi=\frac{1.0\times0.083\times310}{185,000\times0.450}$
$=0.0003$ bar.

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Question 442 Marks

Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.

Answer

A substance dissolves in a solvent if the intermolecular interactions are similar in both the components.
For example: Polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, we can say “like dissolves like”.

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Question 452 Marks

Concentrated nitric acid used in laboratory work is $68\%$ nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $1.504\ g\ mL^{–1}?$

Answer

Concentrated nitric acid used in laboratory work is $68\%$ nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution. Molar mass of nitric acid $(HNO_3) = 1 \times 1 + 1 \times 14 + 3 \times 16 = 63\ g\ mol^{-1}$ Then, number of moles of $\text{HNO}_3=\frac{68}{63}\text{ mol}= 1.079\ mol$
Given,
Density of solution $= 1.504\ g\ mL^{-1}$
Therefore, Volume of $100$ g solution $=\frac{100}{1.504}\text{ mL}$ $= 66.49\ mL = 66.49 \times 10^{-3}\ L$ Molarity of solution $=\frac{1.079\text{ moL}}{66.49\times10^{-3}\text{L}} = 16.23\ M$

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Question 462 Marks

Answer the following question:
Give reasons for the following:

At higher altitudes, people suffer from a disease called anoxia. In this disease, they become weak and cannot think clearly.
When mercuric iodide is added to an aqueous solution of KI, the freezing point is raised.

Answer

At higher altitudes, partial pressure of oxygen is less than that at ground level, so that oxygen concentration becomes less in blood or tissues. Hence, people suffer from anoxia.
Due to the formation of complex $\mathrm{K}_2\left[\mathrm{HgI}_4\right]$, number of particles in the solution decreases and hence the freezing point is raised.

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Question 472 Marks

Vapour pressure of pure water at $298$ K is $23.8$ mm Hg. $50$ g of urea $(NH_2CONH_2)$ is dissolved in $850$ g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer

Vapor pressure of pure water (solvent) at $298\ K, p^0 = 23.8\ mm$
Vapor pressure of solution, $p = ?$
Mass of solvent, $W = 850\ g$
Mass of solute $= 50\ g$
Mol. mass of water $(H_2O),\ M = 18\ g\ mol^{–1}$
Mol. mass of urea $NH_2\ CO\ NH_2$
$= 14 + 2 + 12 + 16 + 14 + 2$
$= 60\ g\ mol^{–1}$
According to Raoult's law,
$=\frac{\text{P}^\circ-\text{P}}{\text{P}^\circ}=\frac{\omega\text{M}}{\text{WM}}$
$\text{P}=\text{P}^\circ-\frac{\omega\text{M}}{\text{Wm}}\times\text{P}^\circ$
$\text{P}=23.8-\frac{50\times18}{60\times850}$
$= 23.8 − 0.017$
$= 23.78$
Hence, $23.78$ mm Hg.

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Question 482 Marks

Calculate the mass percentage of aspirin $\left(\mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_4\right)$ in acetonitrile $\left(\mathrm{CH}_3 \mathrm{CN}\right)$ when 6.5 g of $\mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_4$ is dissolved in 450 g of $\mathrm{CH}_3 \mathrm{CN}$.

Answer

Mass of solute = 6.5 g Mass of solution = 450 + 6.5 = 456.6 g$\text{Mass percentage}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times100$
$=\frac{6.5}{456.5}\times100=\frac{650}{456.5}=1.424\%$

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MCQ 492 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: When methyl alcohol is added to water, boiling point of water increases.
Reason: When a volatile solute is added to a volatile solvent elevation in boiling point is observed.

A
Assertion and reason both are correct statements and reason is correct explanation for assertion.
B
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
C
Assertion is correct statement but reason is wrong statement.
✓
Assertion and reason both are incorrect statements.

Answer

Correct option: D.

Assertion and reason both are incorrect statements.

Methyl alcohol $\ \&\ $ water both are volatile liquids which when mixed together to form a binary solution, the vapour pressure of this solution becomes more as compared to individual pure components. This affects the boiling point of water with a decrease.
When methyl alcohol is added to water $A-B$ interaction $< A-A$ or $B-B$ interaction that is why it will show positive deviation from Raoult’s law. Since a positive deviation from Rault's law indicates a rise in vapour pressure the boiling point decreases.

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Question 502 Marks

Henry’s law constant for the molality of methane in benzene at $298\ K$ is $4.27 \times 10^5\ mm$ Hg. Calculate the solubility of methane in benzene at $298$ K under $760$ mm Hg.

Answer

Here,$p = 760$ mm Hg
$k_H = 4.27 \times 105$ mm Hg
$=$ According to Henry's law,
$P = k_Hx$
$\text{x}=\frac{\text{P}}{\text{k}_\text{H}}$
$=\frac{760\text{ mm Hg}}{4.27\times10^5\text{ mm Hg}}$
$= 177.99 \times 10^{-5}$
$= 178 \times 10^{−5}$ (approximately)
Hence, the mole fraction of methane in benzene is $178 \times 10^{-5}$

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2 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip