Multiple choice

A mixture is prepared by mixing $10$ g $H_{2}SO_{4}$ and $40$ g $SO_{3}$. The mole fraction of $H_{2}SO_{4}$ and $\%$ labelling of oleum is :

  1. $0.169, 118\%$
  2. $0.169, 108\%$
  3. $0.184, 118\%$
  4. $0.184, 108\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

10g H2SO4 (MW 98) is 0.102 mol. 40g SO3 (MW 80) is 0.5 mol. Total moles = 0.602. Mole fraction = 0.102 / 0.602 = 0.169. For oleum labelling, total SO3 is 40g + (10g H2SO4 * 80/98) = 48.16g. Total mass = 50g. % labelling = 100 + (48.16/50 * 18/80 * 100) = 118%.

AI explanation

The moles of H2SO4 are 10 g divided by 98 g/mol, which equals 0.102 moles, and the moles of SO3 are 40 g divided by 80 g/mol, which equals 0.5 moles. The mole fraction of H2SO4 is its moles divided by the total moles (0.602), resulting in 0.169. The percentage of labelling is calculated using the formula 100 + (total mass of SO3 x 80) / (total mass of mixture x 18), which equals 100 + (40 x 80) / (50 x 18), resulting in 118%.

Keep practicing — related questions