Question

The freezing point of pure benzene is 278.4k .Calculate freezing point of the solution when 2gm of a solute having molecular weight 100 is added to 100gm of benzene

Answer 1

Freezing point of the solution when 2gm of a solute having molecular weight 100 is added to 100gm of benzene is 277.4 K

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = $61.82^oC$

$T^o_b$ = freezing point of benzene = $278.4K$

$k_f$ = freezing point constant of benzene = $5.12/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 2 g

$w_1$ = mass of solvent (benzene) = 100 g

$M_2$ = molar mass of solute = 100 g/mol

Now put all the given values in the above formula, we get:

$(278.4-T_f)=1\times (5.12K/m)\times \frac{(2g)\times 1000}{100\times (100g)}$

$T_f=277.4K$

Therefore, the freezing point of solution is 277.4 K

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