Question

The freezing point of a 0.01 M aqueous glucose solution at 1 atmosphere is ${0.18}^{0}C$. To it an addition of equal volume of 0.002 M glucose solution will produce a solution with freezing point of nearly:

• A. ${0.036}^{0}C$
• B. ${0.108}^{0}C$
• C. ${0.216}^{0}C$
• D. ${0.422}^{0}C$

Answer 1

The correct option is B ${0.108}^{0}C$

The freezing point of 0.01 M glucose solution is -0.18 degrees Celsius.

Introducing an equal volume of 0.002 M glucose solution makes the solution with effective molarity of 0.006 M.

Therefore freezing of this new solution, $=-0.18\times 6/10=-{0.108}^{\circ }C$

As $\Delta T_f=T^0-T$

$\therefore -0.108=0-T$

$\therefore T={0.108}^{0}C$

Hence, option B is correct.