Question

A $5$% solution (w/W) of cane sugar (molar mass$=342g$ ${mol}^{-1}$) has freezing point $271K$. What will be the freezing point of $5$% glucose (molar mass $=18g$ ${mol}^{-1}$) in water if freezing point of pure water is $273.15K$?

• A. $273.07K$
• B. $269.07K$
• C. $273.15K$
• D. $260.09K$

Answer 1

The correct option is B $269.07K$

we have,

$\Delta T_f=K_{f}m$

let mass of solution = 100 g

so, weight of canesugar = $\frac{5}{100}\times 100=5g$

and weight of glucose = $\frac{5}{100}\times 100=5g$

so, for cane sugar we have

$273.15-271=K_f\times \frac{5}{342\times 100}1\times 000$

$K_f=14.706$

and for glucose we have,

$\Delta T_f=273.15-T=14.706\times \frac{5}{180\times 100}\times 1000$

hence, freezinf point, $T=269.07K$