Multiple choice

A gaseous mixture contains $CH_4$ and $C_2H_6$ in equimolar proportion. The weight of $1.12\ l$ of this mixture at STP is:

  1. $2.3\ g$
  2. $0.8\ g$
  3. $1.15\ g$
  4. $11.5\ g$
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C Correct answer
Explanation

At STP, 22.4 L of any gas is 1 mole. 1.12 L is 0.05 moles. Mixture is equimolar, so 0.025 moles CH4 and 0.025 moles C2H6. Molar masses: CH4=16, C2H6=30. Mass = 0.025 * 16 + 0.025 * 30 = 0.4 + 0.75 = 1.15 g.

AI explanation

Using the standard molar volume of a gas at STP, 1.12 liters divided by 22.4 liters per mole gives 0.05 moles of the gas mixture. Since the mixture is equimolar, the average molar mass is calculated as (16 g/mol plus 30 g/mol) divided by 2, which equals 23 g/mol. The total mass is found by multiplying the moles by the average molar mass, giving 0.05 moles times 23 g/mol. The final weight of the mixture is therefore 1.15 g.

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