Question

answer the question below​

Answer 1

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▶At 323K, the vapour pressure in millimeters of mercury of methanol - ethanol solution is represented by the equation P = $120\times X_A + 140$, where $X_A$ is the mole fraction of methanol. Then the value of $lim_ x=>1\frac{P_A}{X_A}$

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Given:

Temperature= 323K

P = $120\times X_A + 140$

we have by DALTON'S law of partial pressure!!

$P_A = X_A\times P_T$

where ,

$P_T is total pressure, X_A is mole fraction$

$P_T=\frac{P_A}{X_A}$

let, $P_A = P$

then we have!

$P_T=\frac{P}{X_A}$

$P_T=\frac{120\times X_A + 140}{X_A}$

applying limit!!

we get

$lim_x=>1 \frac{120\times X_A + 140}{X_A}$

$lim_x=>1$$120\times(1) + 140(1)$

$P_T = 120+140=260mm$

Hence , the pressure of mercury on methanol -ethanol is 260mm!