20 g of an non-volatile solute is added to 500 g of solvent, freezing point of pure solvent =5.48° C and that of solution is 4.47° C , K f =1.93 k kg mol -1 molecular mass of solute is:

Question

20 g of an non-volatile solute is added to 500 g of solvent, freezing point of pure solvent =5.48° C and that of solution is 4.47° C , K f =1.93 k kg mol -1 molecular mass of solute is: (A) 77.2 (B) 76.4 (C) 73.2 (D) 70.6

Tardigrade Admin · Accepted Answer

Δ T f =5.48-4.47=1.01, Δ T f = K f × m 1.01=1.93 × (20 × 1000/ m × 50) ⇒ m =76.4

Q. $20 g$ of an non-volatile solute is added to $500 g$ of solvent, freezing point of pure solvent $= 5.4 8^{\circ} C$ and that of solution is $4.4 7^{\circ} C, K_{f} = 1.93 k k g m o l^{- 1}$ molecular mass of solute is:

77.2

76.4

73.2

70.6

$Δ T_{f} = 5.48 - 4.47 = 1.01, Δ T_{f} = K_{f} \times m$

$1.01 = 1.93 \times \frac{20 \times 1000}{m \times 50} \Rightarrow m = 76.4$

Q. 20g of an non-volatile solute is added to 500g of solvent, freezing point of pure solvent =5.48∘C and that of solution is 4.47∘C,Kf​=1.93kkgmol−1 molecular mass of solute is: