Answer to Question #152198 in Physical Chemistry for Ra

Question

The boiling point of chloroform was raised by 0.30K when 5×10⁻⁴ kg of an organic compound was dissolved in 3×10⁻² kg of chloroform. To calculate the molar mass of organic compound. Molal elevation constant for chloroform is 3.9K kg mol⁻¹.

Answer 1

The boiling point elevation can be calculated using the following equation:

$\Delta T = K_bm$ ,

where $\Delta T$ is the change in the temperature, $K_b$ is the molal boiling point elevation constant and $m$ is the molal concentration of the solute in the solution.

From this equation, the molal concentration of the organic compound is:

$m = \frac{\Delta T}{K_b} = \frac{0.30}{3.9} = 0.0769$ mol/kg.

The molality, or the molal concentration is defined as the number of the moles of the solute divided by the mass of the solvent (chloroform is our case):

$m = \frac{n}{m_{solvent}}$ .

Therefore, the number of the moles of the solute present in the solution is:

$n = m·m_{solvent} = 0.0769·0.03 = 0.00231$ mol.

Finally, the molar mass of the solute is:

$M = \frac{m_{solute}}{M} = \frac{5·10^{-4}·1000(g)}{0.00231(mol)} = 216.7$ g/mol.

Answer: the molar mass of the organic compound dissolved in chloroform is 216.7 g/mol.