JEE MAIN - Chemistry (2007 - No. 29)
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour
pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure
(in mm) at the same temperature will be
360
350
300
700
ახსნა
$$P_A^ \circ = ?,\,\,$$
Given $$P_B^ \circ = 200mm,\,\,{x_A} = 0.6$$
$${x_B} = 1 - 0.6 = 0.4,\,\,P = 290$$
$$P = {P_A} + {P_B} = P_A^ \circ {x_A} + P_B^ \circ {x_B}$$
$$ \Rightarrow 290 = P_A^ \circ \times 0.6 + 200 \times 0.4$$
$$\therefore$$ $$\,\,\,\,\,\,\,\,$$ $$P_A^ \circ = 350mm$$
Given $$P_B^ \circ = 200mm,\,\,{x_A} = 0.6$$
$${x_B} = 1 - 0.6 = 0.4,\,\,P = 290$$
$$P = {P_A} + {P_B} = P_A^ \circ {x_A} + P_B^ \circ {x_B}$$
$$ \Rightarrow 290 = P_A^ \circ \times 0.6 + 200 \times 0.4$$
$$\therefore$$ $$\,\,\,\,\,\,\,\,$$ $$P_A^ \circ = 350mm$$