What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10° C ? (kf=1.86 K kg mol -1)

Question

What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10° C ? (kf=1.86 K kg mol -1) (A) 496 g (B) 297 g (C) 310 g (D) 426 g

Tardigrade Admin · Accepted Answer

Δ T f =( K f × W 2 × 1000/ M 2 × W 1) ⇒ W 2=(Δ T f × M 2 × W 1/ k f × 1000) W 2=(10 × 92 × 600/1.86 × 1000)=296.77 g

Q. What weight of glycerol should be added to $600 g$ of water in order to lower its freezing point by $1 0^{\circ} C ?$ $(k_{f} = 1.86 K k g m o l^{- 1})$

496 g

297 g

310 g

426 g

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

$\Rightarrow W_{2} = \frac{Δ T _{f} \times M _{2} \times W _{1}}{k _{f} \times 1000}$

$W_{2} = \frac{10 \times 92 \times 600}{1.86 \times 1000} = 296.77 g$

Q. What weight of glycerol should be added to 600g of water in order to lower its freezing point by 10∘C? (kf​=1.86Kkgmol−1)