Question

a solution containing 18g of non volatile solute in 200g of water freezes at 272.07K.calculate molar mass of the solute

Answer 1

molar mass of the solute is 180
here kf=1.86
change in freezing point is 0.93

Answer 2

Answer : The molar mass of the solute is 155 g/mole.

Solution : Given,

Mass of solute = 18 g

Mass of solvent = 200 g

Initial temperature = 272.07 K

Freezing point of water = 273.15 K (final temperature)

Formula used :

$\Delta T_f=k_f\times m$

where, m is molality

$\Delta T_f=k_f\times m\\(T_{final}-T_{initial})=k_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}\\M_{solute}=\frac{k_f\times w_{solute}}{w_{solvent}\times (T_{final}-T_{initial})}$

where,

$K_f$ = freezing point constant of water = 1.86 K kg/mole

$w_{solute}$ = mass of solute

$w_{solvent}$ = mass of solvent

$M_{solute}$ = molar mass of solute

$T_{initial}$ = initial temperature

$T_{final}$ = final temperature

Now put all the given values in this formula, we get the molar mass of solute.

$M_{solute}=\frac{(1.86Kkg/mole)\times (18g)}{(200g)\times (273.15K-272.07K)}=0.155kg/mole=155g/mole$         $(1kg=1000g)$

Therefore, the molar mass of the solute is 155 g/mole.