Question

Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality of the solution is 0.50 m. (The molar and molal concentrations of dilute aqueous solutions are often identical to two significant figures.) Enter your answer in the provided box. 0.50 m glucose (a covalent compound) °C

Answer 1

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

$T_f-T_f^*=-iK_{solvent}m_{solute}$

Whereas $T_f$ is the freezing temperature of the solution, $T_f^*$ is the freezing temperature of the pure solvent (0 °C since it is water), $i$ the Van't Hoff factor (1 since the solute is covalent), $K_{f,solvent}$ the solvent's freezing point depression point constant (in this case $1.86 C\frac{kg}{mol}$) and $m_{solute}$ the molality of the glucose.

As long as the unknown is $T_f$, solving for it:

$T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg} \\T_f=-0.93C$

Best regards.