By dissolving 5 g substance in 50 g of water, the decrease in freezing point is 1.2° C . The gram molal depression is 1.85° C . The molecular weight of substance is:

Question

By dissolving 5 g substance in 50 g of water, the decrease in freezing point is 1.2° C . The gram molal depression is 1.85° C . The molecular weight of substance is: (A) 105.4 (B) 118.2 (C) 137.2 (D) 154.2

Tardigrade Admin · Accepted Answer

We know that m=(1000 Kf × w/Δ T W) Here, Δ T =1.2° C , Kf=1.85° w =5 g, W=50 g ∴ m =(1000 × 1.85 × 5/1.2 × 50)=154.2

Q. By dissolving $5 g$ substance in $50 g$ of water, the decrease in freezing point is $1. 2^{\circ} C$ . The gram molal depression is $1.8 5^{\circ} C$ . The molecular weight of substance is:

105.4

118.2

137.2

154.2

We know that $m = \frac{1000 K _{f} \times w}{Δ T W}$
Here, $Δ T = 1. 2^{\circ} C, K_{f} = 1.8 5^{\circ}$
$w = 5 g, W = 50 g$
$∴ m = \frac{1000 \times 1.85 \times 5}{1.2 \times 50} = 154.2$

Q. By dissolving 5g substance in 50g of water, the decrease in freezing point is 1.2∘C . The gram molal depression is 1.85∘C . The molecular weight of substance is: