Question

Some non-electrolyte solute (molar mass= 142g/mol) was dissolved in 150 mL of solvent. (density= 0.879 g/cm³). The elevated boiling point of the solution was 355.4K. what mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 kJ/mol, entropy if vaporization is 95.95 J K^-1mol^-1,And the boiling point constant elevation is 2.5 K kg/mol.

Short answer

Mass of solute dissolved in solvent = 15.670 kg

Step-by-step solution

Step 1- Given data

Molar mass of non-electrolyte solute= 142 g/mol

Volume of solvent= 150.mL

Density of solvent= 0.879 g/cm³

Elevated boiling point= 355.4K

Enthalpy of vaporization= 33.90 kJ/mol

Entropy of vaporization = 95.95 J/mol

Step 2- Molar mass

$∆$G= 0 at boiling point, hence

$∆$G= $∆$H – T $∆$S = 0

T= $∆$H/$∆$S

= 33.90 kJ/mol (1000 J/ 1kJ)/ 95.95 J/mol = 353.31 K

$∆$T= K_b m

355.4 – 353.31 K= (2.5 kg. K/mol). m

m= 355.4K – 353.31 K / 2.5 Kg.K /mol = 0.836 mol/kg

Mass of solvent = (150 mL) (0.879 g/ml) (1Kg/1000g) = 0.132 Kg

Mass of solute = (0.132 Kg) (0.836 mol solute/Kg solvent) (142g/1 mol solute)

Mass of solute = 15.670 Kg