Question

A mixture of hydrogen and oxygen has volume 500 \(cm^3\), temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be :

• A. \(3 : 8\)
• B. \(8 : 3\)
• C. \(3 : 16\)
• D. \(16 : 3\)

Hint:

To avoid complex calculations, check the options against the total mass. For option (D), \(16:3\) means \(m_O = \frac{16}{19} \times 0.76 = 0.64\) g and \(m_H = \frac{3}{19} \times 0.76 = 0.12\) g. Quickly calculate moles: \(0.64/32 + 0.12/2 = 0.02 + 0.06 = 0.08\). Since this matches \(PV/RT\), it is correct.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Using the Ideal Gas Equation \(PV = nRT\), we can find the total number of moles in the mixture. Then, using the total mass and individual molar masses, we can find the mass of each component.
Step 2: Key Formula or Approach:
1. \(n_{total} = \frac{PV}{RT}\)
2. \(n_{total} = n_{H_2} + n_{O_2} = \frac{m_{H_2}}{M_{H_2}} + \frac{m_{O_2}}{M_{O_2}}\)
3. Total mass \(m = m_{H_2} + m_{O_2} = 0.76\) g.
Step 3: Detailed Explanation:
Given values:
\(P = 400 \times 10^3\) Pa, \(V = 500 \times 10^{-6}\) \(m^3 = 5 \times 10^{-4}\) \(m^3\), \(T = 300\) K, \(R \approx 8.31\) J/(mol K).
Calculate total moles:
\[ n_{total} = \frac{(4 \times 10^5) (5 \times 10^{-4})}{(8.31) (300)} = \frac{200}{2493} \approx 0.08 \text{ moles} \]
Let \(m_O\) be the mass of oxygen and \(m_H\) be the mass of hydrogen.
Molar mass of \(H_2 = 2\) g/mol, Molar mass of \(O_2 = 32\) g/mol.
Equation 1: \(m_H + m_O = 0.76\)
Equation 2: \(\frac{m_H}{2} + \frac{m_O}{32} = 0.08\)
Multiply Eq 2 by 32:
\[ 16m_H + m_O = 2.56 \]
Subtract Eq 1 from this:
\[ (16m_H + m_O) - (m_H + m_O) = 2.56 - 0.76 \]
\[ 15m_H = 1.80 \implies m_H = \frac{1.80}{15} = 0.12 \text{ g} \]
Find mass of oxygen:
\[ m_O = 0.76 - 0.12 = 0.64 \text{ g} \]
Ratio of masses (Oxygen to Hydrogen):
\[ \frac{m_O}{m_H} = \frac{0.64}{0.12} = \frac{64}{12} = \frac{16}{3} \]
Step 4: Final Answer:
The ratio of the mass of oxygen to hydrogen is \(16 : 3\).