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

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

Step-by-step 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* m`

`\Delta T_b=iK_b* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \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* 5.07^oC kg/mol* (5.65 g)/(M* 0.110 kg)`

`M=1* 5.07^oC kg/mol* (5.65 g)/(1.09^oC* 0.110 kg)`

i = 1

M = 238.91 g/mol

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