Question

What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?

Answer 1

Answer:

$T_f=-2.58\°C$

Explanation:

Hello,

In this case, we can compute the the freezing point depression by using the following formula:

$T_f-T_0=-i*m*Kf$

Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:

$m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m$

Thus, the freezing point of the solution is:

$T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C$

Regards.