Question

Elevation in boiling point was 0.52 when 6 g of a compound x was dissolved in 100 g of water molecular weight of x is

Answer 1

question is incomplete please check the question and retype it

Answer 2

Answer: Molar mass = 59.07 g/mol

Explanation: Elevation in boiling point :

$\Delta T_b=k_b\times m$

where,

$T_b$ = change in boiling point  = 0.52

$k_b$ = elevation constant  = 0.512

m = molality =$\frac{\text{ Moles of solute}}{\text{ Mass of solvent in kg}}$

Moles of solute =$\frac{\text{ given mass}}{\text{molar mass}}$

$\Delta T_f=k_f\times\frac{\text{ given mass}}{\text{molar mass}\times \text{ Mass of solvent in kg}}$

$0.52=0.512\times \frac{6g}{molar mass\times 0.1kg}$

Molar mass = 59.07 g/mol