JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 22)
Liquid 'M' and liquid 'N' form an ideal solution.
The vapour pressures of pure liquids 'M' and
'N' are 450 and 700 mmHg, respectively, at the
same temperature. Then correct statement is:
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
$${{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}$$
(xM – yM) < (xN – yN)
$${{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}$$
$${{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}$$
Explanation
$$P_M^0$$ = 450 mmHg and $$P_N^0$$ = 700 mmHg
$$ \therefore $$ $$P_M^0$$ < $$P_N^0$$
Also we know,
$$P_M$$ = $$P_M^0$$$$X_M$$ = $$Y_M$$$$P_T$$
$$ \Rightarrow $$ $$P_M^0$$ = $${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$$
$$P_N$$ = $$P_N^0$$$$X_N$$ = $$Y_N$$$$P_T$$
$$ \Rightarrow $$ $$P_N^0$$ = $${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$$
As, $$P_M^0$$ < $$P_N^0$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$$ < $${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{X_M}}}$$ < $${{{Y_N}} \over {{X_N}}}$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{Y_N}}}$$ < $${{{X_M}} \over {{X_N}}}$$
$$ \therefore $$ $$P_M^0$$ < $$P_N^0$$
Also we know,
$$P_M$$ = $$P_M^0$$$$X_M$$ = $$Y_M$$$$P_T$$
$$ \Rightarrow $$ $$P_M^0$$ = $${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$$
$$P_N$$ = $$P_N^0$$$$X_N$$ = $$Y_N$$$$P_T$$
$$ \Rightarrow $$ $$P_N^0$$ = $${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$$
As, $$P_M^0$$ < $$P_N^0$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$$ < $${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{X_M}}}$$ < $${{{Y_N}} \over {{X_N}}}$$
$$ \Rightarrow $$ $${{{Y_M}} \over {{Y_N}}}$$ < $${{{X_M}} \over {{X_N}}}$$
Comments (0)
