1

NEET 2021

The correct option for the value of vapour pressure of a solution at 45$^\circ$C with benzene to octane in molar ratio 3 : 2 is :

[At 45$^\circ$C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
A
350 mm of Hg
B
160 mm of Hg
C
168 mm of Hg
D
336 mm of Hg

Explanation

Given : ${n_{{C_6}{H_6}}}:{n_{{C_8}{H_{18}}}} = 3:2$

So, ${\chi _{{C_6}{H_6}}} = {3 \over 5},{\chi _{{C_8}{H_{18}}}} = {2 \over 5}$

Total vapour pressure of solution,

${p_s} = p_{{C_6}{H_6}}^o{\chi _{{C_6}{H_6}}} + p_{{C_8}{H_{18}}}^o{\chi _{{C_8}{H_{18}}}}$

$= 280 \times {3 \over 5} + 420 \times {2 \over 5}$

$= 168 + 168$

$= 336$ mm of Hg
2

NEET 2021

The following solutions were prepared by dissolving 10 g of glucose (C6H12O6) in 250 ml of water (P1), 10 g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is :
A
P3 > P1 > P2
B
P2 > P1 > P3
C
P1 > P2 > P3
D
P2 > P3 > P1

Explanation

$\bullet$ Osmotic pressure ($\pi$) = iCRT where C is molar concentration of the solution

$\bullet$ With increase in molar concentration of solution osmotic pressure increases.

$\bullet$ Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.

$\bullet$ Order of molar mass of solute decreases as Sucrose > Glucose > Urea

$\bullet$ So, correct order of osmotic pressure of solution is P3 > P1 > P2
3

NEET 2020 Phase 1

The mixture which shows positive deviation from Raoult's law is :
A
Benzene + Toluene
B
Acetone + Chloroform
C
Chloroethane + Bromoethane
D
Ethanol + Acetone

Explanation

Pure ethanol molecules are hydrogen bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them. This weakness the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.
4

NEET 2020 Phase 1

The freezing point depression constant (Kr) of benzene is 5. 12 K kg mol-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is rounded off upto two decimal places :
A
0.80 K
B
0.40 K
C
0.60 K
D
0.20 K

Explanation

$\Delta {T_f} = {k_f}m$
= 5.12 $\times$ 0.078
= 0.399 K = 0.4 K