JEE MAIN - Chemistry (2010 - No. 22)
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two
liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of
the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1)
72.0 kPa
36.1 kPa
96.2 kPa
144.5 kPa
Wyjaśnienie
$${P_{Total}} = {P_A}^ \circ {x_A} + {P_B}^ \circ {x_B}$$
$$ = {P^ \circ }_{Hep\tan e}\,{x_{hep\tan e}} + {P^ \circ }_{Oc\tan e}\,{x_{Oc\tan e}}$$
$$ = 105 \times {{25/100} \over {{{25} \over {100}} + {{35} \over {114}}}} + 45 \times {{35/114} \over {{{25} \over {100}} + {{35} \over {114}}}}$$
$$ = 105 \times {{0.25} \over {0.25 + 0.3}} + 45 \times {{0.3} \over {0.25 + 0.3}}$$
$$ = {{105 \times 0.25} \over {0.55}} + {{45 \times 0.3} \over {0.55}}$$
$$ = {{26.25 + 13.5} \over {0.55}} = 72\,kPa$$
$$ = {P^ \circ }_{Hep\tan e}\,{x_{hep\tan e}} + {P^ \circ }_{Oc\tan e}\,{x_{Oc\tan e}}$$
$$ = 105 \times {{25/100} \over {{{25} \over {100}} + {{35} \over {114}}}} + 45 \times {{35/114} \over {{{25} \over {100}} + {{35} \over {114}}}}$$
$$ = 105 \times {{0.25} \over {0.25 + 0.3}} + 45 \times {{0.3} \over {0.25 + 0.3}}$$
$$ = {{105 \times 0.25} \over {0.55}} + {{45 \times 0.3} \over {0.55}}$$
$$ = {{26.25 + 13.5} \over {0.55}} = 72\,kPa$$