Question

A solution is made by dissolving 5.65 g of an unknown molecular compound in 110.0 g of benzene froze at 4.39 oc. what is the molar mass of the solute if pure benzene has a freezing point of 5.45 oc and the kf value of benzene is 5.07 oc/m

Answer 1

Answer is: the molar mass of unknown molecular compound is 245.67 g/mol.

m(compound) = 5.65 g.
m(benzene) = 110 g ÷ 1000 g/kg = 0.11 kg.
ΔT = 5.45°C - 4.39°C.

ΔT = 1.06°C.

Kf(benzene) = 5.07°C/m.

M(compound) = Kf · m(compound) / m(benzene) · ΔT.
M(compound)= 5.07°C/m · 5.65 g / (0.11 kg · 1.06°C).
M(compound) = 245.67 g/mol.

Answer 2

Answer:

238.91 g/mol is the molar mass of the solute.

Explanation:

Mass of solute = 5.65 g

Molar mas of solute = M

Mass of solvent = 110.0 g = 0.110 kg

$\Delta T_f=T-T_f$

$\Delta T_f=K_f\times m$

$\Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$

where,

$\Delta T_f$ =Depression in freezing point

$K_f$ = freezing point constant of solvent= 5.07°C/m

1 =van't Hoff factor

m = molality

Freezing point constant of benzene ,T= 5.45°C/m

T = 5.45°C ,$T_f$ =4.39 °C

$\Delta T_f=T-T_f=5.45^oC-4.39 ^oC=1.09^oC$

$1.09^oC=1\times 5.07^oC kg/mol\times \frac{5.65 g}{M\times 0.110 kg}$

$M=1\times 5.07^oC kg/mol\times \frac{5.65 g}{1.09^oC\times 0.110 kg}$

i = 1

M = 238.91 g/mol

238.91 g/mol is the molar mass of the solute.