Question

The normal boiling point of ethyl acetate is 77.06°C. A solution of 50 g of a non-volatile solute in 150 g of ethyl acetate boils at 84.27°C. Evaluate the molar mass of solute if K_b for ethyl acetate is 2.77°C kg mo1⁻¹.

Answer 1

Given:

T_b = 84.27°C

`T_b^0` = 77.06°C

`\DeltaT_b = T_b - T_b^0`

= 84.27 − 77.06

= 7.21°C

= 7.21 K

W₂ (weight of solute) = 50 g 

W₁ (weight of solvent) = 150 g

K_b (ebullioscopic constant) = 2.77°C kg mol⁻¹

To find:

M₂ (Molar mass of solute) = ?

Formula:

`M_2 = (1000  K_b W_2)/(DeltaT_bW_1)`

Substituting the values in the above formula,

`M_2 = (1000 xx 2.77 xx 50)/(7.21 xx 150)`

M₂ = 128.06 ≅ 128 g mol⁻¹

Molar mass of solute is 128 g mol⁻¹.