Question

one gram of
of a non-volatile solute is dissolved
in 6og benzene. Then freezing point of
benzene
was lowered by 0.4k calculate
the molar mass of the solute if kf of
of benzene
= 1.2 k/m​

Answer 1

Answer:

The molar mass of a non-volatile solute is dissolved  in $60$ g benzene is $50$ g·

Explanation:

Given that,

The mass of the solute,  $W_2$ $=$ $1$ g

The mass of the solvent benzene, $W_1\ =\ 60$ g

The depression in freezing point, $\Delta T_f$  $=$ $0.4$ K

The freezing point depression constant of benzene, $K_f\ =\ 1.2\ \frac{K}{m}$

To calculate the molar mass of the solute, we know that

   $\Delta T_f\ =\ K_f$ × m $---(1)$

Where m denotes the term molality·

We know that

     molality, m $=$ $\frac{No\ of\ mole\ of\ solute}{Mass\ of\ solvent\ in\ Kg}$  $---(2)$

That is,

      m $=$ $\frac{W_2}{M_2}$ × $\frac{1000}{W_1}$

Where,

$W_2\ =$ Mass of solute

$W_1\ =$ Mass of solvent

$M_2\ =$ The molar mass of the solute

On substituting the equation $(2)$ in the equation $(1)$, the equation $(1)$ becomes

   $\Delta T_f\ =\ K_f$ × $\frac{W_2}{M_2}$ × $\frac{1000}{W_1}$

On rearranging the equation with $M_2$ on the left-hand side, the equation becomes

$M_2\ =\ \frac{K_f\ *\ W_2\ *\ 1000}{\Delta T_f\ *\ W_1}$

On substituting the values, we get

 ⇒  $M_2\ =\ \frac{1.2\ *\ 1\ *\ 1000}{0.4\ *\ 60}$

 ⇒  $M_2\ =$ $50$ g

Therefore,

The molar mass of a non-volatile solute is dissolved  in $60$ g benzene is $50$ g·