When 45 g of solute is dissolved in 600 g water, freezing point gets lowered by 2.2 K, calculate molar mass of solute. (Kf = 1.86 K kg mol-1)

Question

When 45 g of solute is dissolved in 600 g water, freezing point gets lowered by 2.2 K, calculate molar mass of solute. (Kf = 1.86 K kg mol-1) (A) 63.4 g mol-1 (B) 80 g mol-1 (C) 90 g mol-1 (D) 21 g mol-1

Tardigrade Admin · Accepted Answer

w1 = 600 g, Δ Tf = 2.2 K w2 = 45 g, Kf = 1.86 K kg mol-1 M = (Kf × w2/Δ Tf × w1)=(1.86×45×1000/2.2×600)=63.4 g mol-1

Q. When 45 g of solute is dissolved in 600 g water, freezing point gets lowered by 2.2 K, calculate molar mass of solute. $(K_{f} = 1.86 K k g m o l^{- 1}$ )

$63.4 g m o l^{- 1}$

$80 g m o l^{- 1}$

$90 g m o l^{- 1}$

$21 g m o l^{- 1}$

$w_{1}$ = 600 g, $Δ T_{f}$ = 2.2 K
$w_{2}$ = 45 g, $K_{f}$ = 1.86 K kg mol $^{- 1}$
M = $\frac{K _{f} \times w _{2}}{Δ T _{f} \times w _{1}} = \frac{1.86 \times 45 \times 1000}{2.2 \times 600} = 63.4 g m o l^{- 1}$

Q. When 45 g of solute is dissolved in 600 g water, freezing point gets lowered by 2.2 K, calculate molar mass of solute. (Kf​=1.86Kkgmol−1)

63.4gmol−1

80gmol−1

90gmol−1

21gmol−1