The depression in freezing point is related to molality of solution as follows:
![\Delta T_{f}=k_{f}m](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%7D%3Dk_%7Bf%7Dm)
Here,
is freezing point depression constant, for cyclohexane it is equal to 20°C kg/mol.
The value of freezing-point depression is 2.5 °C, molality can be calculated as follows:
![m=\frac{\Delta T_{f}}{k_{f}}=\frac{2.5 ^{o}C}{20 ^{o}C kg/mol}=0.125 mol/kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5CDelta%20T_%7Bf%7D%7D%7Bk_%7Bf%7D%7D%3D%5Cfrac%7B2.5%20%5E%7Bo%7DC%7D%7B20%20%5E%7Bo%7DC%20kg%2Fmol%7D%3D0.125%20mol%2Fkg)
Molality is defined as number of moles of solute in 1 kg of solvent.
Here, solvent is cyclohexane, its volume is given 20.0 mL and density is 0.779 g/mol thus, mass of cyclohexane can be calculated as follows:
![m=d\times V=0.779 g/mL\times 20 mL=15.58 g](https://tex.z-dn.net/?f=m%3Dd%5Ctimes%20V%3D0.779%20g%2FmL%5Ctimes%2020%20mL%3D15.58%20g)
Converting this into kg,
![1 g=10^{-3} kg](https://tex.z-dn.net/?f=1%20g%3D10%5E%7B-3%7D%20kg)
Thus, 15.58 g will be 0.01558 kg.
Now, number of moles of unknown solute is related to its mass and molar mass as follows:
![n=\frac{m}{M}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bm%7D%7BM%7D)
Putting the values of mass of solute which is 0.2436 or 0.0002436 kg
![n=\frac{ 0.0002436 kg}{M}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%200.0002436%20kg%7D%7BM%7D)
Now, expression for molality of solution is:
![m=\frac{n_{solute}}{m_{solvent}}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7Bn_%7Bsolute%7D%7D%7Bm_%7Bsolvent%7D%7D)
Putting all the values,
![0.125 mol/kg=\frac{0.0002436 kg}{M\times 0.01558 kg}](https://tex.z-dn.net/?f=0.125%20mol%2Fkg%3D%5Cfrac%7B0.0002436%20kg%7D%7BM%5Ctimes%200.01558%20kg%7D)
Or,
![0.125 mol/kg=\frac{0.015635}{M}](https://tex.z-dn.net/?f=0.125%20mol%2Fkg%3D%5Cfrac%7B0.015635%7D%7BM%7D)
On rearranging,
![M=\frac{0.015635}{0.125 mol/kg}=0.1250 kg/mol](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B0.015635%7D%7B0.125%20mol%2Fkg%7D%3D0.1250%20kg%2Fmol)
Or,
![M=0.125 kg/mol(\frac{1000 g}{1 kg})=125 g/mol](https://tex.z-dn.net/?f=M%3D0.125%20kg%2Fmol%28%5Cfrac%7B1000%20g%7D%7B1%20kg%7D%29%3D125%20g%2Fmol)
Therefore, molar mass of unknown sample is 125 g/mol