Answer:
The correct answer is 16.61 grams methanol and 57.38 grams water.
Explanation:
The mole fraction (X) of methanol can be determined by using the formula,
X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)
X₁ = n₁/n₁ + n₂ = 0.14
n₁ / n₁ + n₂ = 0.14 ---------(i)
n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)
n₁ mole CH₃OH = 32.042 n₁ g
n₂ mole H2O = n₂ mole × 18.015 g/mol
n₂ mole H2O = 18.015 n₂ g
Thus, total mole number is,
32.042 n₁ + 18.015 n₂ = 74 ------------(ii)
From equation (i)
n₁/n₁ + n₂ = 0.14
n₁ = 0.14 n₁ + 0.14 n₂
n₁ - 0.14 n₁ = 0.14 n₂
n₁ = 0.14 n₂ / 1-0.14
n₁ = 0.14 n₂/0.86 ----------(iii)
From eq (ii) and (iii) we get,
32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74
n₂ (32.042 × 0.14/0.86 + 18.015) = 74
n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)
n₂ = 3.1854 mol
From equation (iii),
n₁ = 0.14/0.86 n₂
n₁ = 0.14/0.86 × 3.1854
n₁ = 0.5185 mol
Now, presence of water in the mixture is,
= 3.1854 mole × 18.015 gram per mole
= 57.38 grams
Methanol present in the mixture is,
= 0.5185 mol × 32.042 gram per mole
= 16.61 grams
