Solutions — Page 2 | JEE Chemistry

Q11

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (∆Tf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1)

A 0.0372 K

B 0.0558 K

C 0.0744 K

D 0.0186 K

Correct Answer

Option B

Solution

Sodium sulphate dissociates as

N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - }

hence van't hoff factor

i=3

Now

\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01

= 0.0558\,K

Q12

The degree of dissociation (

\alpha

) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :

A

\alpha = {{i - 1} \over {x + y + 1}}

B

\alpha = {{x + y - 1} \over {i - 1}}

C

\alpha = {{x + y + 1} \over {i - 1}}

D

\alpha = {{i - 1} \over {(x + y - 1)}}

Correct Answer

Option D

Solution

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A_x}\,{B_y}\,\,\rightleftharpoons\,\,{}_x{A^{y + }} + {}_y{B^{x - }}

t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0

{t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\alpha

Total no. of moles

(i)

= 1 - \alpha + x\alpha + y\alpha

i - 1 = x\alpha + y\alpha - \alpha

= \alpha \left( {x + y - 1} \right)

$\therefore$

\,\,\,\,\,\,\,\alpha = {{i - 1} \over {\left( {x + y - 1} \right)}}

Q13

Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8oC ?

A 72 g

B 93 g

C 39 g

D 27 g

Correct Answer

Option B

Solution

\Delta {T_f} = i \times {K_f} \times m

Given

\Delta {T_f} = 2.8,{K_f} = 1.86\,K\,kg\,mo{l^{ - 1}}\,\,i = 1

(ethylene glygol is a non- electrolyte) wt. of solvent

=1

kg;

Let of wt of solute

=x

Mol. wt of ethyllene glycol

=62

2.8 = 1 \times 1.86 \times {x \over {62 \times 1}}

or

\,\,\,\,\,\,x = {{2.8 \times 62} \over {1.86}} = 93gm

Q14

Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

A 0.125 M Na3PO4(aq) has the highest osmotic pressure.

B 0.500 M C2H5OH(aq) has the highest osmotic pressure

C They all have the same osmotic pressure

D 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.

Correct Answer

Option C

Solution

\pi = i\,CRT

^\pi {C_2}{H_3}OH

= 1 \times 0.500 \times R \times T = 0.5{\mkern 1mu} RT

{}^\pi M{g_3}{\left( {P{O_4}} \right)_2}

= 5 \times 0.100 \times R \times T

= 0.5RT

{}^\pi KBr = 2 \times 0.250 \times R \times T = 0.5\,RT

{}^\pi N{a_3}P{O_4} = 4 \times 0.125 \times RT = 0.5\,RT

Since the osmotic pressure of all the given solutions is equal. Hence all are isotonic solution.

Q15

The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :

A 0.0075

B 0.015

C 0.02

D 0.005

Correct Answer

Option B

Solution

Partial pressure = Mole fraction × Solubility

{{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}

$\Rightarrow$

{{500} \over {750}} = {{0.01} \over {{s_2}}}

$\Rightarrow$ s2 = 0.015 g L-1

Q16

5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x (Kf for water=1.86oC kg mol−1) is approximately : (molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)

A 15 g

B 25 g

C 45 g

D 65 g

Correct Answer

Option C

Solution

Na2SO4 $\rightleftharpoons$ 2Na+ + SO4-2 Initial 1 mol 0 0 After dissociation 1 - x 2x x

\therefore\,\,\,

Total no of Moles after dissociation = 1 + 2x Na2SO4 is ionised 81.5% means x = 0.815 Von't Hoff factor (i) =

{{Moles\,\,after\,\,dissociation} \over {Initial\,\,no.\,\,of\,\,moles}}

=

{{1 + 2x} \over 1}

= 1 + 2 $\times$ 0.815 = 2.63

\therefore\,\,\,

\Delta

Tf =

{{1000 \times {K_f} \times {w_2} \times i} \over {{M_S} \times {w_1}}}

$\Rightarrow$

\,\,\,

3.82 =

{{1000 \times 1.86 \times 2.63 \times 5} \over {142 \times x}}

$\Rightarrow$

\,\,\,

x = 45 gm

Q17

The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K on mixing the two liquids, the sum of their initial volume is equal ot the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture, The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :

A 500 mmHg. 0.5,0.5

B 500 mmHg, 0.4, 0.6

C 450 mmHg, 0.4,0.6

D 450 mmHg.0.5,0.5

Correct Answer

Option B

Solution

Given

P_A^0

= 400 mm of Hg

P_B^0

= 600 mm of Hg Mole fraction of B (xB) in liquid phase = 0.5 Mole fraction of A (xA) in liquid phase = 1 - 0.5 = 0.5 Pressure of solution : Ps = xA

P_A^0

+ xB

P_B^0

= 0.5 $\times$ 400 + 0.5 $\times$ 600 = 500 mm of Hg From Roult's law we know, Partial pressure of A (PA) = xA

P_A^0

......(

1) From Dalton'sRoult's law we know, Partial pressure of A (PA) = yAPs ........(2) here yA = mole fraction of A in vapour phase From (1) and (2) we can write, xA

P_A^0

= yAPs $\Rightarrow$ yA =

{{{x_A}P_A^0} \over {{P_s}}}

=

{{0.5 \times 400} \over {500}}

= 0.4 Mole fraction of A in vapour phase = 0.4 $\therefore$ mole fraction of B in vapour phase = 1 - 0.4 = 0.6

Q18

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order : [Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]

A A > C > B

B C > B > A

C A > B > C

D B > C > A

Correct Answer

Option C

Solution

Relative lowering in vapour pressure (RLVP) =

{{P - {P_s}} \over P} = {n \over {n + N}}

n $\to$ moles of solute N $\to$ moles of solvent $\therefore$ (RLVP)A =

{{{{10} \over {100}}} \over {{{10} \over {100}} + {{180} \over {18}}}}

(RLVP)B =

{{{{10} \over {200}}} \over {{{10} \over {200}} + {{180} \over {18}}}}

and (RLVP)C =

{{{{10} \over {10000}}} \over {{{10} \over {10000}} + {{180} \over {18}}}}

$\therefore$ (RLVP)A > (RLVP)B > (RLVP)C So, A > B > C

Q19

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non voltatile solute are together selated in a container. Over time :

A The volume of the solution increases and the volume of the solvent decreases.

B The volume of solution does not change and the volume of the solvent decreases.

C The volumer of the solution and the solvent does not change.

D The volume of the solution decreases and the volume of the solvent increases.

Correct Answer

Option A

Solution

There will be lowering in vapour pressure for solution containing non-volatile solute.

So, there will be transfer of solvent molecules from pure solvent to solution and hence, volume of beaker containing solvent (pure) will decrease and volume of beaker containing solution will increase.

Q20

Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?

A hydrazine

B glucose

C glycine

D KHSO4

Correct Answer

Option D

Solution

ΔTf $\propto$ i × m greater the value of i, greater will be the ΔTf value.

Van't Hoff factor is highest for KHSO4 $\therefore$ colligative property (

\Delta

Tf) will be highest for KHSO4