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JEE Chemistry · 84 questions · Page 7 of 9 · Click an option or "Show Solution" to reveal answer

Q61
A solution containing 62 g ethylene glycol in 250 g water is cooled to −- 10oC. If Kf for water is 1.86 K kg mol−-1 , the amount of water (in g) separated as ice is :
A 48
B 32
C 64
D 16
Correct Answer
Option C
Solution

Here water is solvent and ethylene glycol is solute. We know, Depression of freezing point,

Δ\Delta

Tf = Kf . m

ΔTf=1.86×(6262)(2501000)\Delta {T_f} = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{250} \over {1000}}} \right)}}

= 7.44 We know, freezing point of pure water (here solvent) is 0oC. Which is represented by

Tf0T_f^0

. We also know,

ΔTf=Tf0−Tf\Delta {T_f} = T_f^0 - {T_f}

where

Tf0T_f^0

= Freezing point of pure solvent and

Tf{T_f}

= Freezing point of solution ∴\therefore 7.44 = 0 -

Tf{T_f}

⇒\Rightarrow

Tf{T_f}

= -7.44oC So, not a single drop of water solution will become ice until temperature reaches -7.44oC.

When temperature decrease more than -7.44oC then some part of the solution starts becoming ice staring from surface of the solution.

Now let Wl gm of solution still stays in liquid phase when temperature reaches -10oC.

∴\therefore

10=1.86×(6262)(Wl1000)10 = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{{W_l}} \over {1000}}} \right)}}

Wl = 186 gm So, The amount of water (in g) separated as ice is

Δ\Delta

W = (250 −- 186) = 64 gm

Q62
The solution from the following with highest depression in freezing point/lowest freezing point is
A 180 g180 \mathrm{~g} of acetic acid dissolved in benzene
B 180 g180 \mathrm{~g} of acetic acid dissolved in water
C 180 g180 \mathrm{~g} of benzoic acid dissolved in benzene
D 180 g180 \mathrm{~g} of glucose dissolved in water
Correct Answer
Option B
Solution
ΔTf\Delta \mathrm{T}_{\mathrm{f}}

is maximum when

i×m\mathrm{i} \times \mathrm{m}

is maximum. 1)

m1=18060=3,i=1+α\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha

Hence

ΔTf=(1+α)⋅kf=3×1.86=5.58∘C(α2)\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha 2)

\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}

3)3)

\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=3.8^{\circ} \mathrm{C}

4)4)

\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}{ }^{\prime}=1.86^{\circ} \mathrm{C}

AsperNCERT,As per NCERT,

\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}

\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$$

Q63
Arrange the following solutions in order of their increasing boiling points. (i) 10−4MNaCl10^{-4} \mathrm{M} \mathrm{NaCl} (ii) 10−4M10^{-4} \mathrm{M} Urea (iii) 10−3MNaCl10^{-3} \mathrm{M} \mathrm{NaCl} (iv) 10−2MNaCl10^{-2} \mathrm{M} \mathrm{NaCl}
A (( i )<()<( ii )<()<( iii )<()<( iv ))
B (ii) <(<( i )<()<( iii )<()<( iv ))
C (iv) <(<( iii )<()<( i )<()<( ii ))
D (ii) << (i) ≡\equiv (iii) << (iv)
Correct Answer
Option B
Solution

Step 1: Identify the van’t Hoff factor (ii) for each solute NaCl dissociates (ideally) into two ions: NaCl  →  Na++Cl−, \mathrm{NaCl} \;\rightarrow\; \mathrm{Na^+} + \mathrm{Cl^-}, so i≈2.i \approx 2. Urea (CH4N2O\mathrm{CH_4N_2O}) is a non‐electrolyte (does not dissociate), so i=1.i = 1. Step 2: Effective molar concentration of particles The total particle concentration for each solution is approximately (i×molarity)(i \times \text{molarity}). (i) 10−4 M10^{-4}\,M NaCl Effective concentration  =  2×10−4=2×10−4. \text{Effective concentration} \;=\; 2 \times 10^{-4} = 2 \times 10^{-4}. (ii) 10−4 M10^{-4}\,M Urea Effective concentration  =  1×10−4=1×10−4. \text{Effective concentration} \;=\; 1 \times 10^{-4} = 1 \times 10^{-4}. (iii) 10−3 M10^{-3}\,M NaCl Effective concentration  =  2×10−3=2×10−3. \text{Effective concentration} \;=\; 2 \times 10^{-3} = 2 \times 10^{-3}. (iv) 10−2 M10^{-2}\,M NaCl Effective concentration  =  2×10−2=2×10−2. \text{Effective concentration} \;=\; 2 \times 10^{-2} = 2 \times 10^{-2}. Step 3: Compare to rank the boiling points A larger total particle concentration (and hence larger colligative effect) corresponds to a higher boiling point.

Arrange from lowest to highest: Lowest: 10−4 M10^{-4}\,M Urea [1×10−4]\bigl[1 \times 10^{-4}\bigr] Next: 10−4 M10^{-4}\,M NaCl [2×10−4]\bigl[2 \times 10^{-4}\bigr] Next: 10−3 M10^{-3}\,M NaCl [2×10−3]\bigl[2 \times 10^{-3}\bigr] Highest: 10−2 M10^{-2}\,M NaCl [2×10−2]\bigl[2 \times 10^{-2}\bigr] Hence, in the format (ii)FinalAnswer(\text{ii}) Final Answer \boxed{\text{(ii) }

Q64
Which one of the following 0.06 M aqueous solutions has lowest freezing point?
A Al2(SO4)3
B C6H12O6
C KI
D K2SO4
Correct Answer
Option A
Solution

To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor ii, which represents the number of particles the solute dissociates into when dissolved.

The freezing point depression is given by:

ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m

where: ΔTf\Delta T_f is the change in freezing point, KfK_f is the freezing point depression constant, mm is the molality of the solution, ii is the van 't Hoff factor, which depends on the dissociation of the solute.

For each solute, the van 't Hoff factor ii can be determined as follows: Option A: Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 Dissociates into 2 Al3+^{3+} ions and 3 SO42−_4^{2-} ions.

i=2+3=5i = 2 + 3 = 5 Option B: C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 (glucose) Does not dissociate in solution (non-electrolyte).

i=1i = 1 Option C: KI\text{KI} Dissociates into K+^+ and I−^- ions.

i=1+1=2i = 1 + 1 = 2 Option D: K2SO4\text{K}_2\text{SO}_4 Dissociates into 2 K+^+ ions and 1 SO42−_4^{2-} ion.

i=2+1=3i = 2 + 1 = 3 Comparing the van 't Hoff factors, the solution of Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 will have the highest factor (i=5i = 5), leading to the greatest freezing point depression.

Thus, the Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to ii.

Therefore, the solution of Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 has the lowest freezing point.

Q65
18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :
A 76.0
B 752.4
C 759.0
D 7.6
Correct Answer
Option B
Solution

According to Raoult's Law

P∘−PsPs=WB×MAMB×WA          ...(i){{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{W_B} \times {M_A}} \over {{M_B} \times {W_A}}}\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Here

P∘={P^ \circ } =

Vapour pressure of pure solvent,

Ps={P_s} =

Vapour pressure of solution

WB={W_B} =

Mass of solute,

WA={W_A} =

Mass of solvent

MB={M_B} =

Molar mass of solute,

MA={M_A} =

Molar Mass of solvent Vapour pressure of pure water at

100∘C{100^ \circ }C

(by assumption

=760=760

torr) By substituting values in equation

(i)(i)

we get,

760−PsPs=18×18180×178.2      ...(ii){{760 - {P_s}} \over {{P_s}}} = {{18 \times 18} \over {180 \times 178.2}}\,\,\,\,\,\,...\left( {ii} \right)

On solving

(ii)(ii)

we get On solving

(ii)(ii)

we get

Ps=752.4  torr{P_s} = 752.4\,\,torr
Q66
Freezing point of an aqueous solution is (-0.186)oC. Elevation of boiling point of the same solution is Kb = 0.512 oC, Kf = 1.86 oC, find the increase in boiling point.
A 0.186 oC
B 0.0512 oC
C 0.092 oC
D 0.2732 oC
Correct Answer
Option B
Solution
ΔTb=KbWBMB×WA×1000;\Delta {T_b} = {K_b}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;
ΔTf=KfWBMB×WA×1000;\Delta {T_f} = {K_f}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;
ΔTbΔTf=KbKf=ΔTb−0.186{{\Delta {T_b}} \over {\Delta {T_f}}} = {{{K_b}} \over {{K_f}}} = {{\Delta {T_b}} \over { - 0.186}}
=0.5121.86= {{0.512} \over {1.86}}
=0.0512∘C.= {0.0512^ \circ }C.
Q67
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be. (molar mass of urea = 60 g mol–1)
A 0.031 mmHg
B 0.017 mmHg
C 0.028 mmHg
D 0.027 mmHg
Correct Answer
Option B
Solution

Given that, wsolute = wurea = 0.6 gm wsolvent = wH2O = 360 gm po = 35 We know, lowering of vapour pressure

Δ\Delta

p = xsolute ×\times po =

nureanurea+nH2O{{{n_{urea}}} \over {{n_{urea}} + {n_{{H_2}O}}}}

×\times po =

0.6600.660+36018{{{{0.6} \over {60}}} \over {{{0.6} \over {60}} + {{360} \over {18}}}}

×\times 35 =

10−220×35{{{{10}^{ - 2}}} \over {20}} \times 35

= 0.017

Q68
The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L−10.5 \mathrm{~mL} \mathrm{~L}^{-1} is 0.0405∘C0.0405^{\circ} \mathrm{C}. Density of formic acid is 1.05 g mL−11.05 \mathrm{~g} \mathrm{~mL}^{-1}. The Van't Hoff factor of the formic acid solution is nearly : (Given for water kf=1.86 k kg mol−1\mathrm{k}_{\mathrm{f}}=1.86\, \mathrm{k} \,\mathrm{kg}\,\mathrm{mol}^{-1} )
A 0.8
B 1.1
C 1.9
D 2.4
Correct Answer
Option C
Solution
ΔTf\Delta \mathrm{T}_{\mathrm{f}}

of formic acid

=0.0405∘C=0.0405^{\circ} \mathrm{C}

Concentration

=0.5 mL/L=0.5 \mathrm{~mL} / \mathrm{L}

and density

=1.05 g/mL=1.05 \mathrm{~g} / \mathrm{mL}

∴\therefore Mass of formic acid in solution

=1.05×0.5 g=1.05 \times 0.5 \mathrm{~g}
=0.525 g=0.525 \mathrm{~g}

∴\therefore According to Van't Hoff equation,

ΔTf=ikf⋅m0.0405=i×1.86×0.52546×1\begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned}

(Assuming mass of

1 L1 \mathrm{~L}

water

=kg=\mathrm{kg}

)

i=0.0405×461.86×0.525=1.89≈1.9\mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9
Q69
If α\alpha is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular mass is :
A 1 + α\alpha
B 1 + 2α2\alpha
C 1 - α\alpha
D 1 - 2α2\alpha
Correct Answer
Option B
Solution
Na2 SO4 ⇌2Na+2α  +  SO4−−αN{a_2}\,S{O_4}\,\rightleftharpoons\mathop {2N{a^ + }}\limits_{2\alpha } \,\, + \,\,\mathop {SO_4^{ - - }}\limits_\alpha

Vant. Hoff's factor

   i=1−α+2α+α1=1+2α\,\,\,i = {{1 - \alpha + 2\alpha + \alpha } \over 1} = 1 + 2\alpha
Q70
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
A 0.162
B 0.675
C 0.325
D 0.486
Correct Answer
Option C
Solution

Mole fractions can be calculated as

nCH2Cl2=8.58.5=0.1{n_{C{H_2}C{l_2}}} = {{8.5} \over {8.5}} = 0.1

and

nCHCl3=11.9119.5=0.1{n_{CHC{l_3}}} = {{11.9} \over {119.5}} = 0.1

We know that

ptotal=pCH2Cl2o× xCH2Cl2+pCHCl3o× xCHCl3{p_{total}} = p_{C{H_2}C{l_2}}^o \times \,{x_{C{H_2}C{l_2}}} + p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}
=415×0.10+200×0.1=61.5= 415 \times 0.10 + 200 \times 0.1 = 61.5

Mole fraction of CHCl3 in vapour form can be calculated as

ptotal=pCHCl3o× xCHCl3{p_{total}} = p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}
61.5=200× xCHCl361.5 = 200 \times \,{x_{CHC{l_3}}}
xCHCl3=61.5200=0.3075{x_{CHC{l_3}}} = {{61.5} \over {200}} = 0.3075
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