Question

The freezing point of pure benzene (C₆H₆) is 5. 49 °C. The freezing point of a solution made using toluene (C₇H₈) in benzene is determined to be -13. 0 °C. What is the molality of the toluene? (Kf benzene = 5. 12 °C/m)

Answer 1

The molality of the toluene is 1.46 mol/kg.

What is molality?

The molality of a solution is its molal concentration. Molal concentration is denoted by m. It is the mol of solute dissolves in 1 Kg solvent.

The depression at freezing point is directly proportional to the molality of the solution. The equation we use for this type of problem is:

ΔT=iKm $f_m$

Where,  ΔT $f_m$ is depression at freezing point, i is Van't hoff factor, m is molality and $K_f$ is the freezing point depression constant.

Toluene is a non-electrolyte and so the value of i for this is 1.

Depression at a freezing point is given as -13.$0^0$ and the $K_f$ is given as 5.12. So, we could calculate the molality of the solution using the equation written above.

Let's plug in the values in the equation:

ΔT=iKm$f_m$

-7.51 =1 x5. 12 xm

m = $\frac{7.51}{5. 12}$

m = 1.46 m

It means the molality of the solution is 1.46 mol/kg.

Hence, the molality of the solution is 1.46 mol/kg.

Learn more about the molality of the solution here:

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