3 Marks Question

Question 513 Marks

Vapour pressure of water at $293$ K is $17.535$ mm Hg. Calculate the vapour pressure of water at $293$ K when $25$ g of glucose is dissolved in $450$ g of water.

Answer

Vapour pressure of water, $\text{P}^0_1=17.535\text{ mm of Hg}$
Mass of glucose, $w_2 = 25$ g Mass of water, $w_1 = 450$ g
We know that,
Molar mass of glucose $(C_6H_{12}O_6), M_2 = 6 \times 12 + 12 \times 1 + 6 \times 16$
$= 180\ g\ mol^{−1}$ Molar mass of water, $M_1 = 18\ g\ mol^{-1}$
Then, number of moles of glucose, $\text{n}_2=\frac{25}{180\text{ g mol}^{-1}}= 0.139$ mol And,
number of moles of water, $\text{n}_1=\frac{450}{18\text{ g mol}^{-1}} = 25$ mol
We know that, $\frac{\text{P}^0_1-\text{P}_1}{\text{P}^0_1}=\frac{\text{n}_1}{\text{n}_2+\text{n}_1}$
$\frac{17.535-\text{P}_1}{17.535}=\frac{0.139}{0.139+25}$
$17.535-\text{P}_1=\frac{0.139\times17.535}{25.139}$
$17.535-\text{P}_1=0.097$
$p_1 = 17.44$ mm of Hg
Hence, the vapour pressure of water is $17.44$ mm of Hg.

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Question 523 Marks

Determine the osmotic pressure of a solution prepared by dissolving $25$ mg of $K_2SO_4$ in $2$ litre of water at $25^\circ C$, assuming that it is completely dissociated.

Answer

When $K_2S0_4$ is dissolved in water, $K^+$ and $\text{SO}^{2-}_4$ ion produced$\text{K}_2\text{SO}_4\rightarrow2\text{K}^++\text{SO}^{2-}_4$
total number of ion produced 3
therfore, $i = 3$
given that
$w = 25, mg= 0.0259$
$T = 25^0C + 273 = 298\ K$
Also we know that
$R = 0.0821\ L\ atm\ K^{-1}\ mol^{-1}$
$M = (2 \times 39) + (1 \times 32) + (4 \times 16)$
Osmotic pressure, $\pi=?$
$V = 2L,\ i = 3$
$\pi=\frac{\text{i}\text{ n}_2\text{ RT}}{\text{V}}=\frac{\text{i w RT}}{\text{m V}}$
$\pi=\frac{3\times25\times10^{-3}\times0.0821\times298}{174\times2}$
$= 5.27 \times 10^{-3}$ atm.

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Question 533 Marks

Calculate the mass of a non-volatile solute $($molar mass $40$ g mol$^{–1})$ which should be dissolved in $114$ g octane to reduce its vapour pressure to $80\%$.

Answer

$P_s = 80\%$ of $P^0=\frac{80}{100}\text{ P}^\circ=0.8\text{ P}^\circ$
Let $W_g$ of solute is present in mixture.
Moles of solute present $=\frac{\text{W}}{40}\text{ moles}$
Molar mass of octane, $C_8H_{18}$
$= 8 x 12 + 18 = 114\ gmol^{-1}$
$\therefore\ \text{Moles of octane}=\frac{114}{114}=1\text{ mol}$
$\text{Now},\ \frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\text{x}_2=\frac{\frac{\text{W}}{40}}{\frac{\text{W}}{40}+1}$
$\frac{\text{P}^\circ-0.80\text{ P}^\circ}{\text{P}^\circ}=\frac{\frac{\text{W}}{40}}{\frac{\text{W}}{40}+1}$
$1-0.80=\frac{\text{W}\times40}{40(\text{W}+40)}=\frac{\text{W}}{\text{W}+40}$
$0.20=\frac{\text{W}}{\text{W}+40}$
$0.2 W + 8 = W$
$8 = W(l - 0.2)$
$8 = 0.8\ W$
$\therefore\ \text{W}=\frac{8}{0.8}=10g.$

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Question 543 Marks

A solution containing $30$ g of non-volatile solute exactly in $90$ g of water has a vapour pressure of $2.8$ kPa at $298$ K. Further, $18$ g of water is then added to the solution and the new vapour pressure becomes $2.9$ kPa at $298$ K. Calculate:

Molar mass of the solute.
Vapour pressure of water at $298$ K.

Answer

Let the molar mass of solute $= Mg\ mol^{-1}$
$\therefore\ \text{Moles of solute present}$ $=\frac{30\text{g}}{\text{Mg mol}^{-1}}=\frac{30}{\text{M}}\text{ mol}$
$\text{Moles of solvent present}, (\text{n}_1)=\frac{90}{18}=5\text{ moles.}$
$\therefore\ \frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\frac{\text{n}_2}{\text{n}_1+\text{n}_2}$
$\frac{\text{P}^\circ-2.8}{\text{P}^\circ}=\frac{\frac{30}{\text{M}}}{\frac{5+30}{\text{M}}}$
$1-\frac{2.8}{\text{P}^\circ}=\frac{30}{(5\text{M}+30)}$
$1-\frac{30}{5\text{M}+30}=\frac{2.8}{\text{P}^\circ}$
$1-\frac{6}{\text{M}+6}=\frac{2.8}{\text{P}^\circ}$
$\frac{\text{M}+6-6}{\text{M}+6}=\frac{2.8}{\text{P}^\circ}$
$\frac{\text{M}}{\text{M}+6}=\frac{2.8}{\text{P}^\circ}$
$\frac{\text{P}^\circ}{2.8}=1+\frac{6}{\text{M}}\ ....(\text{i})$
After adding 18 g of water,
Moles of water becomes $=\frac{90+18}{18}=\frac{108}{18}=6\text{ moles}$
$\therefore\ \frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\frac{\frac{30}{\text{M}}}{\frac{6+30}{\text{M}}}$
$P_s$ New vapour pressure = 2.9 KPa .
$\frac{\text{P}^\circ-2.9}{\text{P}^\circ}=\frac{30\text{M}}{\text{M}(6\text{M}+30)}=\frac{5}{\text{M}+5}$
$1-\frac{2.9}{\text{P}^\circ}=\frac{5}{\text{M}+5}$
$1-\frac{5}{\text{M}+5}=\frac{2.9}{\text{P}^\circ}$
$\frac{\text{M}+5-5}{\text{M}+5}=\frac{2.9}{\text{P}^\circ}$
$\frac{\text{P}^\circ}{2.9}=\frac{\text{M}+5}{\text{M}}\Rightarrow\ =1+\frac{5}{\text{M}}$
$\frac{\text{P}^\circ}{2.9}=1+\frac{5}{\text{M}}\ .... (\text{ii})$ Dividing equation (i) by (ii),
we get, $\frac{2.9}{2.8}=\frac{1+\frac{6}{\text{M}}}{1+\frac{5}{\text{M}}}$
$2.9\Big(1+\frac{5}{\text{M}}\Big)=2.8\Big(1+\frac{6}{\text{M}}\Big)$
$2.9+\frac{2.9\times5}{\text{M}}=2.8+\frac{2.8\times6}{\text{M}}$
$2.9+\frac{14.5}{\text{M}}=2.8+\frac{16.8}{\text{M}}$
$0.1=\frac{16.8}{\text{M}}-\frac{14.5}{\text{M}}=\frac{2.3}{\text{M}}$
$\text{M}=\frac{2.3}{0.1}$
$M = 23\ g\ mol^{-1}$ Putting $M = 23,$ in equation (i), we get,
$\frac{\text{P}^\circ}{2.8}=1+\frac{6}{23}=\frac{29}{23}$
$\text{P}^\circ=\frac{29}{23}\times2.8=3.53\text{ KPa}$

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Question 553 Marks

An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004$ bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer

Vapour pressure of pure water at the boiling poin$(\text{P}^\circ)=1.013\text{ bar}$
Vapour pressure of solution $(Ps) = 1·004$ bar
Mass of solute $(w_2) = 2g$ Molar
mass of solvent, water $(M_1) = 18g$ Mass of solvent $(w_1) = 98g$
Mass of solution $= 100g$
Applying Raoult's Law for dilute solutions, $\frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\frac{\text{n}_2}{\text{n}_1+\text{n}_2}\simeq\frac{\text{n}_2}{\text{n}_1}$
$[$Dilute solution being $2\%]$ $\frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\frac{\text{n}_2}{\text{n}_1}=\frac{\frac{\text{W}_2}{\text{M}_2}}{\frac{\text{W}_1}{\text{M}_1}}$
$\frac{(1.013-1.004)}{(1.013)}=\frac{2\times18}{\text{M}_2\times98}$
$\therefore\ \text{M}_2=\frac{2\times18}{98\times0.009}\times1.013=41.35\text{ g mol}^{-1}.$

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Question 563 Marks

Calculate the molarity of each of the following solutions:

$30$ g of $Co(NO_3)_2. 6H_2O$ in $4.3$ L of solution.
$30$ mL of $0.5$ $M H_2SO_4$ diluted to $500$ mL.

Answer

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

Mol. mass of $Co (NO_3). 6H_2O$

$= 58.9 + (14 + 3 × 16)2 + 6(18)$
$= 58.9 + (14 + 48) × 2 + 108$
$= 58.9 + 124 + 108 = 290.9$
Moles of $Co (NO)_3. 6H_2O$
$=\frac{30}{290.9}=0.103\text{ mol}$
Volume of solution $= 4.3$ L
Molarity,
$\text{M}=\frac{\text{Moles of solute}}{\text{Volume of solution in litre}}$
$=\frac{103}{4.3}=0.024\text{ M}.$

Number of moles present in $1000$ ml of $0.5$ M $H_2SO_4= 0.5$ mol

therefore number of moles present in $30$ml of $0.5$M $H_2SO_4=\frac{0.5\times30}{1000}\text{ mol}$
$= 0.015$ mol
Therefore molarity $= 0.015/0.5$ L
Thus molarity is $0.03$M.

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Question 573 Marks

Explain the following:

Solution of chloroform and acetone is an example of maximum boiling azeotrope.
A doctor advised a person suffering from high blood pressure to take less quantity of common salt.

Answer

This solution has lesser vapour pressure due to stronger interactions (hydrogen bonds) between chloroform and acetone molecules.
Because higher quantity of NaCl will increase number of sodium and chloride ions in the body fluid which can increase the osmotic pressure of body fluid, i.e., blood pressure of a person.

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Question 583 Marks

Boiling point of water at $750$ mm Hg is $99.63^\circ C.$ How much sucrose is to be added to $500$ g of water such that it boils at $100^\circ C.$

Answer

Here, elevation of boiling point $\triangle T_b = (100 + 273) - (99.63 + 273) = 0.37 K$
Mass of water, $w_1 = 500 $g Molar
mass of sucrose $(C_{12}H_{22}O_{11}), M_2 = 11 \times 12 + 22 \times 1 + 11 \times 16 = 342$ g mol
Molal elevation constant, $K_b = 0.52$ K kg mol$^{-1}$
We know that:
$\Delta\text{T}_\text{b}=\frac{\text{K}_\text{b}\times1000\times\text{w}_2}{\text{M}_2\times\text{w}_1}$ $\Rightarrow\text{w}_2=\frac{\Delta\text{T}_\text{b}\times\text{M}_2\times\text{w}_1}{\text{K}_\text{b}\times1000}$$=\frac{0.37\times342\times500}{0.52\times1000}$
$= 121.67$ g (approximately) Hence, $121.67$ g of sucrose is to be added.

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Question 593 Marks

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at $300 K$ are $50.71$ mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if $80 g$ of benzene is mixed with $100 g$ of naphthalene.

Answer

Molar mass of benzene $\left(\mathrm{C}_6 \mathrm{H}_6\right)=6 \times 12+6 \times 1=78 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of toluene $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3\right)=7 \times 12+8 \times 1=$ $92 \mathrm{~g} \mathrm{~mol}^{-1}$ Now, no. of moles present in 80 g of benzene $=\frac{80}{78} \mathrm{~mol}=1.026 \mathrm{~mol}$ And, no. of moles present in 100 g of toluene $=\frac{100}{92} \mathrm{~mol}$
$=1.087 \mathrm{~mol}$ Therefore, Mole fraction of benzene, $\mathrm{x}_{\mathrm{b}}=\frac{1.026}{1.026+1.087}=0.486$ And, mole fraction of toluene, $\mathrm{x}_{\mathrm{t}}=1$ $0.486=0.514$ It is given that vapour pressure of pure benzene, $\mathrm{P}_{\mathrm{b}}^0=50.71 \mathrm{~mm} \mathrm{Hg}$ And, vapour pressure of pure toluene, $\mathrm{P}_1^0=32.06 \mathrm{~mm}$ HG Therefore, partial vapour pressure of benzene, $\mathrm{P}_{\mathrm{b}}=\mathrm{x}_{\mathrm{b}} \times \mathrm{P}_{\mathrm{b}}=0.486 \times 50.71=24.645$ mm Hg And, partial vapour pressure of toluene, $P_1=x_t \times P_t=0.514 \times 32.06=16.479 \mathrm{~mm} \mathrm{Hg}$ Hence, mole fraction of benzene in vapour phase is given by: $\frac{\mathbf{P}_{\mathrm{b}}}{\mathbf{P}_{\mathrm{b}}+\mathbf{P}_{\mathrm{t}}}$
$=\frac{24.645}{24.645+16.479}$
$=\frac{24.645}{41.124}$
$=0.599=0.6$

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Question 603 Marks

Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
$NaCl(s)$ and $H_2O(l)$

Answer

For a solution containing non-volatile solute i.e. $NaCl(s)$ and $H_2O(l),$ the Raoult’s law is applicable only to vaporisable component (1) i.e. $H_2O(l)$ and total vapour pressure is written as,
$\text{p}=\text{p}_1=\text{x}_1\text{p}^0_1$
Where ${p_1}^o$ reresents the vapour pressure of pure $H_2O(l)$

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Question 613 Marks

An antifreeze solution is prepared from $222.6 g$ of ethylene glycol $(C_2H_6O_2)$ and $200 g$ of water. Calculate the molality of the solution. If the density of the solution is $1.072 g mL^{–1}$, then what shall be the molarity of the solution?

Answer

Molar mass of ethylene glycol $\left[\mathrm{C}_2 \mathrm{H}_4(\mathrm{OH})_2\right]=2 \times 12+6 \times 1+2 \times 16=62 \mathrm{~g} \mathrm{~mol}^{-1}$
Number of moles of ethylene glycol $\frac{222.6 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}$
$=3.59 \mathrm{~mol}$
Therefore, molality of the solution $=\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}}$
$=17.95 \mathrm{~m}$
Total mass of the solution $=(222.6+200) \mathrm{g}$
$=422.6 \mathrm{~g}$
Given,
Density of the solution $=1.072 \mathrm{~g} \mathrm{~mL}^{-1}$
Therefore, Volume of the solution $=\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}$
$=394.22 \mathrm{~mL}$
$=0.3942 \times 10^{-3} \mathrm{~L}$
Molarity of the solution $=\frac{3.59 \mathrm{~mol}}{0.39422 \times 10^{-3} \mathrm{~L}}$
$=9.11 \mathrm{M}$

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Question 623 Marks

A solution of glucose in water is labelled as $10\%$ w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is $1.2\ g\ mL^{–1},$ then what shall be the molarity of the solution?

Answer

10% w/w solution of glucose in water means that $10$ g of glucose in present in $100$ g of the solution
i.e., $10$ g of glucose is present in $(100 - 10)\ g = 90\ g$ of water.
Molar mass of glucose $(C_6H_{12}O_6) = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180\ g\ mol^{-1}$ Then,
number of moles of glucose $=\frac{10}{180}\text{ mol} = 0.056$ mol
$\text{Molality of solution}=\frac{0.056\text{ mol}}{0.09\text{ kg}}=0.62\text{m}$ $\text{Number of moles of water}=\frac{90\text{g}}{18\text{ g mol}^{-1}}$= 5 mol
Mole fraction of glucose $(x_g) =\frac{0.056}{0.056+5}$
$= 0.011$ And, mole fraction of water $x_w= 1 - x_g = 1 - 0.011 = 0.989$
If the density of the solution is $1.2g\ ml^{−1} $, then the volume of the $100$ g solution can be given as:
$=\frac{100\text{g}}{1.2\text{ g ml}^{-1}}= 83.33\ mL = 83.33 \times 10^{-3}L$
$\therefore\ \text{Molarity of the solution}=\frac{0.056\text{ mol}}{83.33\times10^{-3}\text{L}} = 0.67\ M$

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Question 633 Marks

When fruits and vegetables that have dried are placed in water, they slowly swell and return to the original form. Explain why. Would a temperature increase accelerate the process? Explain.

Answer

The cell walls of the fruits and vegetables act as semipermeable membrane. When they are dried, concentration inside becomes higher. On placing in water, the process of osmosis takes place. So, they swell and return to their original form. The process will be accelerated with increase of temperature because osmosis becomes faster with increase in temperature.

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Chemistry 3 Marks Question