Question

Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1oC. What is the molar mass of this substance?

Answer 1

Raoult's law states:

$T_b - T_s = KC_m$, where T_b -- freezing point of benzene (5.5^oC), T_s -- freezing point of the solution, K -- cryoscopic constant (5.12 K*kg/mol for benzene), C_m -- molality of solution.

$C_m = m_l/Mm_s$ , where m_l -- mass of lauryl alcohol, m_s -- mass of the solvent.

Combine the two formulas:

$T_b - T_s = Km_l/Mm_s$

Find the molar mass:

$M = Km_l/m_s(T_b-T_s)$

$M = 5.12*5/0.1(5.5-4.1) =$ 183 (g/mol)

Answer: 183 g/mol -- the molar mass of lauryl alcohol. (The real one is 186 g/mol).