An aqueous solution containing 5% by weight of urea and 10% by weight of glucose. What will be its freezing point ( $K_{f} = 1.86 K k g m o l^{-}$ ). (write only integer part)

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Solution

Urea =5g; Glucose =10 g
Weight of solute = 15 g
Weight of solvent =100-15 =85 g
$m_{u r e a} = m_{1} = \frac{W_{2} \times 1000}{M W_{2} \times W_{1}} = \frac{5 \times 1000}{60 \times 85} = 0.98$
$m_{g l u c o s e} = m_{2} = \frac{10 \times 1000}{180 \times 85} = 0.653$
$Δ T_{f} = K_{f} (m_{1} + m_{2}) = 1.86 (0.98 + 0.653) = 3.038$
Freezing point = $273.15 - 3.038 = 270.11 K$

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