2 Marks Questions

Question 512 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 522 Marks

Calculate the mass of ascorbic acid $($Vitamin $C, C_6H_8O_6)$ to be dissolved in $75 g$ of acetic acid to lower its melting point by $1.5^\circ C. Kf = 3.9 K \ kg \ mol^{-1}.$

Answer

$\text{Given:} \ \Delta\text{T}_\text{f}=1.5^\circ$ Mass of $CH_3COOH, w_1 = 75g M_1 = 60g \ mol^{-1}M_2(C_6H_8O_6) = 176g \ mol^{-1}$
$K_f = 3.9K \ Kg mol^{-1}$
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 532 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 $CO_2$ above the solution decreases suddenly, $($as per Henry's law$).$This results into a decrease in solubility of carbon dioxide, hence $CO_2$ bubbles come out of the bottle with a fizz.

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Question 542 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 552 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, $V = 450\ ml = 0.450L$
Temperature, $T = (37 + 273)K = 310K \pi=?, \text{V}=450\text{ mL}=0.450\text{L},$
$T = 273 + 37 = 310K \ R = 0.083 L \ bar \ mol^{-1}K^{-1}, M_2 = 185,000, W_2 = 1.0g$
$\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 562 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 572 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 $\text{HNO}_3=\frac{68}{63}\text{ mol}= 1.079 \ mo$l
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 582 Marks

Calculate the mass percentage of aspirin $(C_9H_8O_4)$ in acetonitrile $(CH_3CN)$ when $6.5 g$ of $C_9H_8O_4$ is dissolved in $450 g$ of $CH_3CN.$

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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Question 592 Marks

If the solubility product of CuS is $6 \times 10^{–16},$ calculate the maximum molarity of $CuS$ in aqueous solution.

Answer

Solubility product of $CuS, K_{sp} = 6 x 10^{-16}$
$\text{CuS}\rightleftharpoons\text{Cu}^{2+}+\text{S}^{2-}$
Suppose solubility of $CuS$ is $x mol^{-1}$ This would give $\times mol^{-1} of Cu^{2+} $ions and $\times mol L^{-1} of S^{2-}$
ions on dissociation. $[Cu^{2+}] = x mol^{-1} [S^{2-}] = x mol^{-1}$
$\text{K}_\text{sp}=[\text{Cu}^{2+}][\text{S}^{2-}]=(\text{x})(\text{x})=\text{x}^2$
$\text{K}_\text{sp}=6\times10^{-16}$
$\therefore\ 6\times10^{-16}=\text{x}^2$
$\text{or}\ \text{ x}=\sqrt{6\times10^{-16}}=2.45\times10^{-8}$
Maximum molarity $= 2.45 x 10^{-8} M.$

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Question 602 Marks

Answer the following question:
Why a person suffering from high blood pressure is advised to take minimum quantity of common salt?

Answer

Osmotic pressure is directly proportional to the concentration of the solutes. Our body fluid contains a number of solutes. On taking large amount of common salt, $Na^+$ and $Cl^-$ ions enter into the body fluid thereby raising the concentration of the solutes. As a result, osmotic pressure increases which may rupture the blood cells.

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Question 612 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 $\ce{NH_2 CO NH_2}$
$= 14 + 2 + 12 + 16 + 14 + 2$
$= 60\ g\ mol^{–1}$
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 622 Marks

What do you understand by azeotropic mixture Define minimum azeotropic mixture and give an example.

Answer

Azeotropic Mixture: It is a liquid mixture that boils at a constant temperature and has the same composition in both the liquid and vapor phases. It cannot be separated by simple distillation.
Minimum Boiling Azeotrope: A mixture that shows a large positive deviation from Raoult's Law. Its boiling point is lower than that of either of its components.
→ Example: 95% Ethanol + 5% Water (by volume).

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Chemistry 2 Marks Questions