Relative Molecular Mass And Mole

We have, K_f = 1.86 K kg mol^{– 1} 
w = 0.2 g,
W = 21.7 g,
ΔT_f= 272.814 K.

 Osmotic Pressure. The pressure which is applied to the solution in order to prevent the passage of solvent into it through a semi-permeable membrane present between the two is termed as osmotic pressure.

We know that osmotic pressure is a colligative property and hence it depends on the number of solute particles.

Since    1 mole = 6.023 x 10²³ particles

So    Sucrose = 6.023 x 10²² particles

(As Sucrose and Glucose are undissociated)

Glucose = 30.115 x 10²² particles

NaCl = 12.046 x 10²² particles.

(NaCl dissociates to give two ions)

Hence, the increasing order of their osmotic pressure is, Sucrose < NaCl < Glucose.

B.

373.0512K

[0-(-0.186)] = 1.86 x m

$$\begin{gathered} \therefore \mathrm{m} = \frac{0.186}{1.86} = 0.1 \\ ∆{\mathrm{T}}_{\mathrm{b}} = {\mathrm{K}}_{\mathrm{b}}.\mathrm{m} \\ \therefore {\mathrm{T}}_{\mathrm{b}} = 0.512 \mathrm{x} 0.1 \\ = 0.0512 \\ \mathrm{T}- {\mathrm{T}}^{\mathrm{o}} = 0.0512 \\ \mathrm{T} = 0.0512 + {\mathrm{T}}^{\mathrm{o}} \\ = 0.0512 + 373 \\ = 373.0512 \mathrm{K} \end{gathered}$$

As given,

Mass of K₂SO₄ = 0.025 gm

Molar mass of K₂SO₄ = 174 g/mol

Volume = 2L

Temperature  = 25^oC (298 K)

$$\begin{gathered} \mathrm{\pi } = \frac{\mathrm{n}}{\mathrm{V}}\mathrm{RT} \\ = \frac{\mathrm{mass}}{\mathrm{Molar} \mathrm{mass}} \mathrm{x} \frac{1}{\mathrm{V}}\mathrm{RT} \\ \frac{0.025}{174} \mathrm{x} \frac{1}{2} \mathrm{x} 8.314 \mathrm{x} 298 \end{gathered}$$

 = 0.173 atm