Question

Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 percent by weight, has a freezing point
Equal to -0.89 ° C, set Kf value of water = 1.86 ° C / m

Answer 1

given

7% by weight

freezing point = -0.89⁰ C

Kf value of water = 1.86 ° C / m

we know that

Tf = I Kf m

where

T_f is freezing point depression

which is T_{f (pure solvent)} − T_{f (solution)} and T_{f (pure solvent)} is 0⁰C

K_f = 1.86 ° C / m

m is molality

I is the van 't Hoff factor

so

Tf = T_{f (pure solvent)} − T_{f (solution)}

==> Tf = (0⁰C) - (-0.89⁰C) = 0.89⁰C

so

0.89⁰C = I Kb m

since substance is not specified let I = 1

0.89⁰C = 1.86 ° C / m * m

molality = 0.47849 m

now

7% by weight

which means 7 g of substance in 100 g of water

so the solution has 7 g of substance and

100 g - 7 g of solvent = 93 g solvent

molality = no of moles solute / mass of solvent in kg

= (mass of solute/molecular weight of solute) / mass of solvent in kg

0.47849 m = ( 7 g / molecular weight of solute) / (93 * 10^(-3) kg)

hence

molecular weight of solute = ( 7 g / ( 0.47849 mol/kg * 0.093 kg) ) = 157.3 g/mol