Question

3. The vapour pressures of ethyl alcohol and methyl alcohol are 45 mm Hg and 90 mm Hg. An ideal
solution is formed at the same temperature by mixing 60 g of C2H5OH with 40 g of CH3OH. Total
vapour pressure of the solution is approximately -
(A) 70 mm
(B) 35 mm
(C) 105 mm
(D) 27 mm Hg
​

Answer 1

HLO MATE HERE U GO

⭐. OPTION A is correct

Answer 2

Answer:

Total pressure of the solution is equal to $67.01mmHg$.

Therefore option (A) is correct.

Explanation: Molar mass of ethyl alcohol $=42g$

Weight of ethyl alcohol $=60g$

Number of moles of ethyl alcohol $n_{A}=\frac{60}{42}=1.304mol$

Molar mass of methanol $=32g$

Weight of  methanol $=40g$

Number of moles of  methanol $n_{B}=\frac{40}{32}=1.25mol$

Mole fraction of $C_{2}H_{5}OH$, $x_{A}=\frac{1.304}{1.304+1.25} =0.5107$

Mole fraction of$CH_{3}OH$ $x_{B}=\frac{1.25}{1.304+1.25} =0.4893$

Partial pressure of $C_{2}H_{5}OH$, $x_{A}p^{o}_{A}=(0.5107)(45)=22.981mmHg$

Partial pressure of $CH_{3}OH$, $x_{B}p^{o}_{B}=(0.4893)(90)=44.037mmHg$

Total pressure of solution, $P=x_{A}p^{o}_{A}+x_{B}p^{o}_{B}$

 ∴        $P=44.037+22.981\\P=67.018mmHg$

Therefore total pressure of solution is equal to $67.018mmHg$.