Question

1 g of a non- electrolyte solute(molar mass 25g/mol) was dissolved in 51.2 g of benzene. If the freezing point depression constant , K_f of benzene is 5.12 K kg /mol , the freezing point of benzene will be lowerd by

1) 0.2K

2) 0.4K

3) 0.3K

4) 0.5K

Answer 1

Solve by using formula ΔT_f = K_f $\times$ i$\times$ m (1)
Where , ΔT_{f is the} change in freezing point of pure benzene and solution of non-electrolyte in benzene
K_f is molal freezing point depression constant for benzene = 5.12 Kkg/mol
i is the dissociated species of solute, i=1 as solute is a non-electrolyte
m is molality of solute = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
$\text{m}=\frac{m_{solute}}{{\text{M}}_{\text{solute}}{\text{× m}}_{\text{solvent}}\text{in kg}}$
m_solute, mass of solute = 1g
M_solute, molar mass of solute = 25gmol^-1
m_solvent, mass of solvent benzene = 51.2g = 0.0512 kg

Equation 1 can be modified to ΔT_f = $\frac{{\text{K}}_f\times i\times m_{solute}}{{\text{M}}_{solute}\times m_{solvent}}$
Substituting values
ΔT_f = $\frac{5.12\times 1\times 1}{25\times 0.0512}$

ΔT_f = 4K
Change in freezing point or freezing point of benzene is lowered by 4K