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221. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol^–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of gluocose (mol. mass = 180 g mol^–1) per 100 g of solution?

Answer:

The molality of the cane sugar, m= 0.1539m
depression in freezing point $∆ T_{f}$ = 273.15-271
=2.15K

Since $∆ T_{f}$ = K_fm
or K_f = $∆ T_{f}$ /m = 2.15k/0.1539m = 13.97K/m

Now the weight of glucose , W₂ = 5g
molecular mass of glucose, M₂ =180g/mol

then $∆ T_{f}$ =K_fm

$∆ T_{f}$ =

$\frac{K_{f} \times W_{2} \times 1000}{W_{1} \times M_{2}} = \frac{13.97 \times 5 \times 1000}{100 \times 180} = 3.88 K$

then freezing point of solution = 273.15-3.88
=263.27K

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221. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of gluocose (mol. mass = 180 g mol–1) per 100 g of solution?

221. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol^–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of gluocose (mol. mass = 180 g mol^–1) per 100 g of solution?