The freezing point of a solution, prepared from 1.25 gm. of a non-electrolyte and 20 gm. of water, is 271.9 K. If the molar depression constant is 1.86 K / mol, then molar mass of the solute is

Question

The freezing point of a solution, prepared from 1.25 gm. of a non-electrolyte and 20 gm. of water, is 271.9 K. If the molar depression constant is 1.86 K / mol, then molar mass of the solute is (A) 103.3 (B) 105.7 (C) 107.9 (D) 109.3

Tardigrade Admin · Accepted Answer

The molar mass of solute =(1000 × Kf × W/Δ Tf × w) Here weight of solute (W)=1.25 g weight of solvent (w)=20 g Molar depression constant (Kf)=1.86 Freezing point of a solution (Δ Tf) =(273-271.9) =1.1 K Putting the values, the molar mass of the solute =(1000 × 1.86 × 1.25/1.1 × 20)=105.7

Q. The freezing point of a solution, prepared from 1.25gm. of a non-electrolyte and 20gm. of water, is 271.9K. If the molar depression constant is 1.86K/mol, then molar mass of the solute is

103.3

105.7

107.9

109.3

The molar mass of solute =ΔTf​×w1000×Kf​×W​ Here weight of solute (W)=1.25g weight of solvent (w)=20g Molar depression constant (Kf​)=1.86 Freezing point of a solution (ΔTf​)=(273−271.9) =1.1K Putting the values, the molar mass of the solute =1.1×201000×1.86×1.25​=105.7

Q. The freezing point of a solution, prepared from $1.25 g m$ . of a non-electrolyte and $20 g m$ . of water, is $271.9 K$ . If the molar depression constant is $1.86 K / m o l$ , then molar mass of the solute is