Answered

The molecular weight of an organic compound was determined by measuring the freezing point depression of a benzene solution. A 0.500 g sample was dissolved in 50.0 g of benzene, and the resulting depression was 0.42°C (kf (benzene) = 5.065°C/m). What is the approximate molecular weight of the compound?

Answer :

Answer: The molar mass of the sample is 120.6 g/mol

Explanation:

To calculate the depression in freezing point, we use the equation:

[tex]\Delta T_f=iK_fm[/tex]

Or,

[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]

where,

[tex]\Delta T_f[/tex] = depression in freezing point = 0.42°C

i = Vant hoff factor = 1 (For non-electrolytes)

[tex]K_f[/tex] = molal freezing point elevation constant = 5.065°C/m

[tex]m_{solute}[/tex] = Given mass of solute (sample) = 0.500 g

[tex]M_{solute}[/tex] = Molar mass of solute (sample) = ? g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (benzene) = 50.0 g

Putting values in above equation, we get:

[tex]0.42^oC=1\times 5.065^oC/m\times \frac{0.500\times 1000}{M_{solute}\times 50.0}\\\\M_{solute}=\frac{1\times 5.065\times 0.500\times 1000}{50.0\times 0.42}=120.6g/mol[/tex]

Hence, the molar mass of the sample is 120.6 g/mol

Other Questions