Solutions — Page 4 | JEE Chemistry

Q31

At 35oC, the vapour pressure of CS2 is 512 mm. Hg and that of acetone is 344 mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is :

A CS2 and acetone are less attracted to each other than to themselves.

B a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml.

C Heat must be absorved in order to produce the solution 35oC

D Raoult's law is not obeyed by this system.

Correct Answer

Option B

Solution

As the vapour pressure of mixture increases so the force of attraction of mixture is less than the individual component.

Since the vapour pressure of the solution is greater than individual vapour pressure of both pure components, the solution shows a positive deviation from Raoult’s law.

$\therefore$

\Delta

SolH

>

0 As

\Delta

SolH

>

0 then solution is endothermic. and here

\Delta

Sol Volume

>

0. As

\Delta

Sol Volume

>

0 then a mixture of 100 ml.

CS2 and 100 ml. acetone has a volume more than 200 ml.

It happens because the distance between molecules increases due to force of attraction of mixture decreases.

Q32

What weight of glucose must be dissolved in

100 \mathrm{~g}

of water to lower the vapour pressure by

0.20 \mathrm{~mm} ~\mathrm{Hg}

? (Assume dilute solution is being formed) Given : Vapour pressure of pure water is

54.2 \mathrm{~mm} ~\mathrm{Hg}

at room temperature. Molar mass of glucose is

180 \mathrm{~g} \mathrm{~mol}^{-1}

A 3.69 g

B 2.59 g

C 3.59 g

D 4.69 g

Correct Answer

Option A

Solution

The lowering of vapor pressure of a solvent by a nonvolatile solute is given by Raoult's law:

\Delta P = x_{\text{solute}} \cdot P_0

where $\Delta P$ is the change in vapor pressure, $x_{\text{solute}}$ is the mole fraction of the solute, and $P_0$ is the vapor pressure of the pure solvent.

Rearranging the formula for $x_{\text{solute}}$ , we have:

x_{\text{solute}} = \frac{\Delta P}{P_0}

Substituting the given values, we get :

x_{\text{solute}} = \frac{0.20 \, \text{mm Hg}}{54.2 \, \text{mm Hg}} = 0.003689

Since the mole fraction of the solute is also equal to the number of moles of solute divided by the total number of moles, we can express $x_{\text{solute}}$ as:

x_{\text{solute}} = \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{solute}} + \text{moles}_{\text{water}}}

Assuming that the solution is dilute, the number of moles of water will be much larger than the number of moles of solute, so we can approximate the total number of moles as the number of moles of water.

Thus, we have :

x_{\text{solute}} \approx \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{water}}}

Therefore, the number of moles of solute is :

\text{moles}_{\text{solute}} = x_{\text{solute}} \cdot \text{moles}_{\text{water}}

The number of moles of water is the mass of the water divided by the molar mass of water (18 g/mol) :

\text{moles}_{\text{water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}

So, the number of moles of solute is :

\text{moles}_{\text{solute}} = 0.003689 \cdot 5.56 \, \text{mol} = 0.0205 \, \text{mol}

Finally, to find the mass of the glucose, we multiply the number of moles of glucose by the molar mass of glucose :

\text{mass}_{\text{glucose}} = \text{moles}_{\text{solute}} \cdot \text{molar mass}_{\text{glucose}}

\text{mass}_{\text{glucose}} = 0.0205 \, \text{mol} \cdot 180 \, \text{g/mol} = 3.69 \, \text{g}

So, the correct answer is 3.69 g.

Q33

Equimolar solutions in the same solvent have

A Same boiling point but different freezing point

B Same freezing point but different boiling point

C Same boiling and same freezing points

D Different boiling and different freezing points

Correct Answer

Option C

Solution

Equimolar solutions of normal solutes in the same solvent will have the same boiling point and same freezing point.

Q34

Molal depression constant for a solvent is 4.0 kg mol–1. The depression in the freezing point of the solvent for 0.03 mol kg–1 solution of K2SO4 is : (Assume complete dissociation of the electrolyte)

A 0.18 K

B 0.24 K

C 0.36 K

D 0.12 K

Correct Answer

Option C

Solution

K2SO4 $\to$ 2K+ + SO42- Van’t Hoff Factor (i) = 3 $\therefore$

\Delta

Tf = ikfm = 3 $\times$ 4 $\times$ 0.03 = 0.36 K

Q35

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is : (R = 0.08206 L atm K–1 mol–1)

A 8.2 atm

B 2.46 atm

C 4.92 atm

D 1.64 atm

Correct Answer

Option C

Solution

We know, osmotic pressure ( $\pi$ ) = (Resultant molarity) $\times$ R $\times$ T Resultant molarity =

{{Total\,moles\,of\,all\,solutions} \over {Total\,volume(in\,lit)}}

=

{{{{0.6} \over {60}} + {{1.8} \over {180}}} \over {0.1}}

= 0.2 mol/lit $\therefore$ Osmotic pressure ( $\pi$ ) = (0.2) $\times$ 0.08206 $\times$ 300 = 4.92 atm

Q36

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6oC will be : [Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol = 62 g mol−1 )

A 204.30 g

B 400.00 g

C 304.60 g

D 804.32 g

Correct Answer

Option D

Solution

Given

{K_r} = 1.86\,K\,kg\,mo{l^{ - 1}}

\Delta {T_f} = 0 - \left( { - 6} \right) = {6^ \circ }C

As we know that

\Delta {T_f} = {K_f} \times \,\,molality

= {{{K_f} \times 1000 \times mass\,\,of\,\,solute} \over {molar\,\,mass\,\,of\,\,solute\,\, \times \,\,mass\,\,of\,\,solvent\,\,in\,\,kg}}

Substituting given values in formula

6 = {{1.86 \times 1000 \times w} \over {62 \times 4}};

\,\,\,\,\,w = 0.8\,kg = 800\,gm

Q37

A 5.25 % solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol−1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm−3, molar mass of the substance will be

A 90.0 g mol−1

B 115.0 g mol−1

C 105.0 g mol−1

D 210.0 g mol−1

Correct Answer

Option D

Solution

TIPS/FORMULAE: Osmotic pressure

\left( \pi \right)

of isotonic solutions are equal. For solution of unknown substance

\left( {\pi = CRT} \right)

{C_1} = {{5.25/M} \over V}

For solution of urea,

C{}_2\,\,

(concentration)

= {{1.5/60} \over V}

Given,

{\pi _1} = {\pi _2}

as

\,\,\,\,\pi = CRT

$\therefore$

\,\,\,\,\,{C_1}RT = {C_2}RT\,\,

or

\,\,{C_1} = {C_2}

or

\,\,\,\,{{5.25/M} \over V} = {{1.8/60} \over V}

$\therefore$

\,\,\,\,\,M = 210g/mol

Q38

Evaluate the following statements for their correctness. A. The elevation in boiling point temperature of water will be same for

0.1 \mathrm{M} \, \mathrm{NaCl}

and

0.1 \mathrm{M}

urea. B. Azeotropic mixtures boil without change in their composition. C. Osmosis always takes place from hypertonic to hypotonic solution. D. The density of

32 \% \, \mathrm{H}_{2} \mathrm{SO}_{4}

solution having molarity

4.09 ~\mathrm{M}

is approximately

1.26 \mathrm{~g} \mathrm{~mL}^{-1}

E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution. Choose the correct answer from the options given below :

A A, B and D only

B A and C only

C B and D only

D B, D and E only

Correct Answer

Option C

Solution

(A) Elevation in boiling point temperature of water will be higher for 0.1 M NaCl as compared to 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition (C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.

(E) When KI solution is added to AgNO3 solution, positively charged solution is formed due to adsorption of Ag+ ions from dispersion medium.

AgI/ Ag+ (positively charged) (D) Let the mass of H2SO4 (32%) is 100.

\therefore \text{ Weight of } \mathrm{H}_2 \mathrm{SO}_4=32

Moles of $\mathrm{H}_2 \mathrm{SO}_4=\dfrac{32}{98}$ Now, $4.09=\dfrac{32}{98 \times \mathrm{V}} \Rightarrow \mathrm{V}=79 \mathrm{ml}$ Density $=\dfrac{100}{79}=1.265$ Hence, correct answer is (C) B and D only.

Q39

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water

=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

)

A 377.3 K

B 379.2 K

C 375.3 K

D 277.3 K

Correct Answer

Option A

Solution

Boiling Point Elevation Formula: $\Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b$ where: $i_1$ and $i_2$ are the van't Hoff factors for ethylene glycol and glucose, respectively (both are 1 since they do not dissociate in solution). $m_1$ and $m_2$ are the molalities of ethylene glycol and glucose, respectively. $K_b$ is the ebullioscopic constant of water ( $0.52 \, \text{K kg mol}^{-1}$ ).

Calculate Molality: Each solute has 2 moles dissolved in 500 grams of water ( $0.5 \, \text{kg}$ ): $m_1 = \dfrac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1}$ $m_2 = \dfrac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1}$ Substitute into the Formula: $\Delta T_b = 1 \cdot 4 \cdot 0.52 + 1 \cdot 4 \cdot 0.52 = 4.16$ Determine Boiling Point of Solution: The normal boiling point of water is $373.16 \, \text{K}$ .

Add the boiling point elevation to the normal boiling point: $T_b (\text{solution}) = 373.16 + 4.16 = 377.3 \, \text{K}$ Thus, the boiling point of the resulting solution is $377.3 \, \text{K}$ .

Q40

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

A [Co(H2O)3 Cl3].3H2O

B [Co(H2O)6] Cl3

C [Co(H2O)5 Cl] Cl2.H2O

D [Co(H2O)4 Cl2] Cl.2H2O

Correct Answer

Option A

Solution

We know,

\Delta {T_f}\, = \,im{K_f}

Where

\Delta {T_f}\, = \,

Depression in freezing point i = no of ions or molecule m = molality of solute Kf = freezing point depression constant.

In this question m = 1 and Kf is constant.

So,

\Delta {T_f}\,\, \propto \,\,i

Which solutions have more i, those solutions

\Delta {T_f}\,

will be more. Now, Freezing point of a aqueous solution

{f_p} = \, - \,\Delta {T_f}

So, those solutions have more

\Delta {T_f}

will have less freezing point and

\Delta {T_f}

will be more when value of i is more.

Now, for, [ Co (H2O)6] Cl3 $\rightleftharpoons$ [ Co(H2O)6]3+ + 3Cl $-$ Value of i = 4 (no. of ions), 3 ions of Cl $-$ and one ion of [ Co (H2O)6]3+ [ Co (H2O)5 Cl ] Cl2 .

H2O $\rightleftharpoons$ [ Co (H2O)5 Cl ]2+ + 2Cl $-$ Value of i = 3 [ Co (H2O)4 Cl2 ] Cl .

2H2O $\rightleftharpoons$ [ Co (H2O)4 Cl2 ] + + Cl $-$ Value of i = 2 and [ Co(H2O)3 Cl ] .

3H2O do not show ionization.

So, i = 1.

As, [ Co (H2O)3 Cl ].

3H2O have lowest value of i, So, it's freezing point will be maximum.