Solutions — Page 3 | NEET Chemistry

Q21

Which of them is not equal to zero for an ideal solution?

A

\Delta {V_{mix}}

B

\Delta P = {P_{observed}} - {P_{Raoult}}

C

\Delta {H_{mix}}

D

\Delta {S_{mix}}

Correct Answer

Option D

Solution

For an ideal solution,

\Delta

S mix > 0 while

\Delta

H mix ,

\Delta

V mix and

\Delta

P all are 0.

Q22

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

A KCl

B C 6 H 12 O 6

C Al 2 (SO 4 ) 3

D K 2 SO 4

Correct Answer

Option C

Solution

We know that depression in freezing point (

\Delta

T f ) is given as

\Delta

T f = iK f m So,

\Delta

T f $\propto$ i Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.

Al 2 (SO 4 ) 3 ⇌ 2Al +3 + 3SO 4 2– i is maximum i.e., 5 for Al 2 (SO 4 ) 3 . .

Q23

Which condition is not satisfied by an ideal solution?

A

{\Delta _{mix}}\,V = 0

B

{\Delta _{mix}}\,S = 0

C Obeyance to Raoult's Law

D

{\Delta _{mix}}\,H = 0

Correct Answer

Option B

Solution

An ideal solution is follow: 1. Volume change (

\Delta

V) of mixing should be zero 2. Heat change (

\Delta

H) on mixing should be zero. 3. Obey Raoult’s law at every range of concentration.

Q24

Vapour pressure of chloroform (CHCl 3 ) and dichloromethane (CH 2 Cl 2 ) at 25 o C are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl 3 and 40 g of CH 2 Cl 2 at the same temperature will be (Molecular mass of CHCl 3 = 119.5 u and molecular mass of CH 2 Cl 2 = 85 u)

A 173.9 mm Hg

B 615.0 mm Hg

C 90.6 mm Hg

D 285.5 mm Hg

Correct Answer

Option C

Solution

n CHCl 3 =

{{25.5} \over {119.5}}

= 0.213 n CH 2 Cl 2 =

{{40} \over {85}}

= 0.47 P T = P o A X A + P o B X B =

200 \times {{0.213} \over {0.683}} + 41.5 \times {{0.47} \over {0.683}}

= 62 + 28.55 = 90.63

Q25

p A and p B are the vapour pressure of pure liquid components, A and B , respectively of an ideal binary solution. If x A represents the mole fraction of component A , the total pressure of the solution will be

A

{p_A} + {x_A}\left( {{p_B} - {p_A}} \right)

B

{p_A} + {x_A}\left( {{p_A} - {p_B}} \right)

C

{p_B} + {x_A}\left( {{p_B} - {p_A}} \right)

D

{p_B} + {x_A}\left( {{p_A} - {p_B}} \right)

Correct Answer

Option D

Solution

p = p A x A + p B x B = p A x A + p B (1 – x A ) (As for binary solution x A + x B = 1) = p A x A + p B – p B x A = p B + x A (p A – p B )

Q26

200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57

\times

10

-

3 bar. The molar mass of protein will be (R = 0.083 L bar mol

-

1 K

-

1 )

A 51022 g mol

-

1

B 122044 g mol

-

1

C 31011 g mol

-

1

D 61038 g mol

-

1

Correct Answer

Option D

Solution

Osmotic pressure, $\pi$ = CRT $\Rightarrow$ $\pi$ =

{n \over V}

RT $\Rightarrow$ $\pi$ V =

{w \over M}

RT M =

{{wRT} \over {\pi V}}

=

{{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times {{200} \over {1000}}}}

=

{{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times 0.2}}

= 61038 g mol –1

Q27

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K f for water is 1.86 o C/m, the freezing point of the solution will be

A

-

0.18 o C

B

-

0.54 o C

C

-

0.36 o C

D

-

0.24 o C

Correct Answer

Option D

Solution

We know that

\Delta

T f = i × K f × m Here i is van’t Hoff’s factor. i for weak acid is 1 + $\alpha$ .

Here $\alpha$ is degree of dissociation i.e., 30/100 = 0.3 $\therefore$ i = 1 + $\alpha$ = 1 + 0.3 = 1.3 $\therefore$ T f = i × K f × m = 1.3 × 1.86 × 0.1 = 0.24 $\therefore$ Freezing point = – 0.24

Q28

The freezing point depression constant for water is

-

1.86 o C m

-

1 . If 5.00 g Na 2 SO 4 is dissolved in 45.0 g H 2 O, the freezing point is changed by

-

3.82 o C. Calculate the van't Hoff factor for Na 2 SO 4 .

A 2.05

B 2.63

C 3.11

D 0.381

Correct Answer

Option B

Solution

According to depression in freezing point,

\Delta

T f = i $\times$ K f $\times$ m = i $\times$ K f $\times$

{{{w_B} \times 1000} \over {{m_B} \times {w_A}}}

Given :

\Delta

T f = 3.82, K f = 1.86, w B = 5, m B = 142, w A = 45 i =

{{\Delta T \times {m_B} \times {w_A}} \over {{K_f} \times {w_B} \times 1000}}

=

{{3.82 \times 142 \times 45} \over {1.86 \times 5 \times 1000}}

= 2.63

Q29

The van't Hoff factor

i

for a compound which undergoes dissociation in one solvent and association in other solvent is respectively

A less than one and greater than one

B less than one and less than one

C greater than one and less than one

D greater than one and greater than one.

Correct Answer

Option C

Solution

From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association.

In case of dissociation, i is greater than 1 and in case of association i is less than 1.

Q30

A solution of sucrose (molar mass = 342 g mol

-

) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ( K f for water = 1.86 K kg mol

-

1 )

A

-

0.372 o C

B

-

0.520 o C

C + 0.372 o C

D

-

0.570 o C

Correct Answer

Option A

Solution

Depression in freezing point,

\Delta

T f = K f $\times$ m m =

{{{w_B}} \over {{M_B}}} \times {{1000} \over {{W_A}}}

=

{{68.5 \times 1000} \over {342 \times 1000}}

\Delta

T f = 1.86 $\times$

{{68.5} \over {342}}

= 0.372 o C $\therefore$ T f = 0 - 0.372 o C = - 0.372 o C