A cooking gas contains carbon and hydrogen only. A volume of 11.2 L of this gas is found to weigh 22 g at STP. Then the molecular formula of the gas is
$C_2H_6$
- $C_2H_2$
- $C_2H_4$
$C_3H_8$
- $C_3H_4$
The Correct Option is D
Solution and Explanation
We are given:
- Volume of gas = 11.2 L
- Mass of gas = 22 g
- Conditions = STP (Standard Temperature and Pressure)
At STP, 1 mole of any ideal gas occupies 22.4 L. So,
Moles of gas = \( \frac{11.2}{22.4} = 0.5 \)
Molar mass = \( \frac{22}{0.5} = 44 \, \text{g/mol} \)
We are told the gas contains only carbon (C) and hydrogen (H). Let’s test molecular masses:
- \( C_3H_8 \): \( 3 \times 12 + 8 \times 1 = 44 \, \text{g/mol} \) ✅
- \( C_2H_2 \): \( 24 + 2 = 26 \, \text{g/mol} \)
- \( C_2H_4 \): \( 24 + 4 = 28 \, \text{g/mol} \)
- \( C_2H_6 \): \( 24 + 6 = 30 \, \text{g/mol} \)
- \( C_3H_4 \): \( 36 + 4 = 40 \, \text{g/mol} \)
Only \( C_3H_8 \) has the correct molar mass.
Correct Answer: (D) \( C_3H_8 \)
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