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Q. What weight of glycerol should be added to $600\, g$ of water in order to lower its freezing point by $10 \,{}^°C$ ?
$(K_f = 1.86^°\, C\, m^{-1})$

Solutions

Solution:

$\Delta T_{f} = K_{f} \times m = \frac{K_{f} \times W_{2} \times 1000}{M_{2} \times W_{1}}$
$W_{2} = \frac{10 \times 92 \times 600}{1.86 \times 1000} =296.77\,g$