Question:
A vessel of volume \(20\,\text{L}\) contains a mixture of hydrogen and helium at temperature \(27^\circ\text{C}\) and pressure \(2\,\text{atm}\). The mass of the mixture is \(5\,\text{g}\). Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the mixture is:
A vessel of volume \(20\,\text{L}\) contains a mixture of hydrogen and helium at temperature \(27^\circ\text{C}\) and pressure \(2\,\text{atm}\). The mass of the mixture is \(5\,\text{g}\). Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the mixture is:
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For gas mixtures:
\[
n=\sum \frac{m}{M}
\]
Use molecular masses carefully.
Updated On: Mar 23, 2026
- \(1:2\)
- \(2:3\)
- \(2:1\)
- \(2:5\)
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Total moles: \[ n=\frac{PV}{RT} \]
Step 2: Let masses be \(m_1\) (H\(_2\)) and \(m_2\) (He): \[ \frac{m_1}{2} + \frac{m_2}{4} = n \]
Step 3: Given: \[ m_1+m_2=5 \]
Step 4: Solving gives: \[ m_1:m_2 = 2:1 \]
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