JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 8)
1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose
ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, $${{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}$$, is :
5 : 1
1 : 0.2
10 : 1
1 : 5
Explanation
Given,
$${{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} = {1 \over 5}$$
$$\Delta {T_b}\left( A \right) = {k_b}\left( A \right) \times $$ Molality of A
$$\Delta {T_b}\left( B \right) = {k_b}\left( B \right) \times $$ Molality of B
Here, Molality of A = Molality of B = m
$${{\Delta {T_b}\left( A \right)} \over {\Delta {T_b}\left( B \right)}} = {{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} \times {m \over m}$$ = $${1 \over 5}$$
$${{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} = {1 \over 5}$$
$$\Delta {T_b}\left( A \right) = {k_b}\left( A \right) \times $$ Molality of A
$$\Delta {T_b}\left( B \right) = {k_b}\left( B \right) \times $$ Molality of B
Here, Molality of A = Molality of B = m
$${{\Delta {T_b}\left( A \right)} \over {\Delta {T_b}\left( B \right)}} = {{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} \times {m \over m}$$ = $${1 \over 5}$$
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