2.0 g of a non-electrolyte dissolved in 100 g of benzene lowers the freezing point of benzene by 1.2 K. The freezing point depression constant of benzene is 5.12 K kg mol -1. The molar mass of the solute is

Question

2.0 g of a non-electrolyte dissolved in 100 g of benzene lowers the freezing point of benzene by 1.2 K. The freezing point depression constant of benzene is 5.12 K kg mol -1. The molar mass of the solute is (A) 55 g mol -1 (B) 85 g mol -1 (C) 120 g mol -1 (D) 155 g mol -1

Tardigrade Admin · Accepted Answer

Δ Tf=Kf m=Kf × (W2/M2) × (1000/w1) 1 ⋅ 2=(5 ⋅ 12 × 2 × 1000/M2 × 100) M2=85 g mol -1

Q. $2.0 g$ of a non-electrolyte dissolved in $100 g$ of benzene lowers the freezing point of benzene by $1.2 K$ . The freezing point depression constant of benzene is $5.12 K k g m o l^{- 1}$ . The molar mass of the solute is

$55 g m o l^{- 1}$

$85 g m o l^{- 1}$

$120 g m o l^{- 1}$

$155 g m o l^{- 1}$

$Δ T_{f} = K_{f} m = K_{f} \times \frac{W _{2}}{M _{2}} \times \frac{1000}{w _{1}}$

$1 \cdot 2 = \frac{5 \cdot 12 \times 2 \times 1000}{M _{2} \times 100}$

$M_{2} = 85 g m o l^{- 1}$

Q. 2.0g of a non-electrolyte dissolved in 100g of benzene lowers the freezing point of benzene by 1.2K. The freezing point depression constant of benzene is 5.12Kkgmol−1. The molar mass of the solute is

55gmol−1

85gmol−1

120gmol−1

155gmol−1