JEE MAIN - Chemistry (2020 - 6th September Evening Slot - No. 15)
A set of solutions is prepared using 180 g of
water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative
lowering of vapour pressure in the presence of
these solutes are in the order :
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
A > C > B
C > B > A
A > B > C
B > C > A
Explanation
Relative lowering in vapour pressure (RLVP)
= $${{P - {P_s}} \over P} = {n \over {n + N}}$$
n $$ \to $$ moles of solute
N $$ \to $$ moles of solvent
$$ \therefore $$ (RLVP)A = $${{{{10} \over {100}}} \over {{{10} \over {100}} + {{180} \over {18}}}}$$
(RLVP)B = $${{{{10} \over {200}}} \over {{{10} \over {200}} + {{180} \over {18}}}}$$
and (RLVP)C = $${{{{10} \over {10000}}} \over {{{10} \over {10000}} + {{180} \over {18}}}}$$
$$ \therefore $$ (RLVP)A > (RLVP)B > (RLVP)C
So, A > B > C
= $${{P - {P_s}} \over P} = {n \over {n + N}}$$
n $$ \to $$ moles of solute
N $$ \to $$ moles of solvent
$$ \therefore $$ (RLVP)A = $${{{{10} \over {100}}} \over {{{10} \over {100}} + {{180} \over {18}}}}$$
(RLVP)B = $${{{{10} \over {200}}} \over {{{10} \over {200}} + {{180} \over {18}}}}$$
and (RLVP)C = $${{{{10} \over {10000}}} \over {{{10} \over {10000}} + {{180} \over {18}}}}$$
$$ \therefore $$ (RLVP)A > (RLVP)B > (RLVP)C
So, A > B > C
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