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° C m-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° C m-1) (A) 496 g (B) 297 g (C) 310 g (D) 426 g

Tardigrade Admin · Accepted Answer

Δ Tf = Kf × m = (Kf × W2 × 1000/M2 × W1) W2 = (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 $10^{°} C$ ?
$(K_{f} = 1.8 6^{°} C m^{- 1})$

$496 g$

$297 g$

$310 g$

$426 g$

$Δ T_{f} = K_{f} \times m = \frac{K _{f} \times W _{2} \times 1000}{M _{2} \times W _{1}}$
$W_{2} = \frac{10 \times 92 \times 600}{1.86 \times 1000} = 296.77 g$

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

496g

297g

310g

426g