Find the depression in freezing point of a solution containing 0.520 g glucose $(C_{6} H_{12} O_{6})$ dissolved in 80.2 g of water. [ Given, $K_{f}$ for water=1.86 $K m^{- 1}$

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Solution

According to the problem:
Let us consider the molecular mass of glucose
$C_{6} H_{12} O_{6} = 72 + 12 + 96$
Implies that
$= 180 g m o l^{- 1}$
Hence the freezing point,
$Δ T_{f} = \frac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}$
=>
$= \frac{1.86 \times 0.520 \times 1000}{180 \times 80.20}$
$= 0.067$

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