In a 74.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.140. What is the mass of each component?

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

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;