Question

A mixture of hydrogen and oxygen has volume 2000 \(cm^3\), temperature 300 K, pressure 100 kPa and mass 0.76 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:[Take gas constant R = 8.3 \(JK^{–1}\) \(mol^{–1}\)]

• A. \(\frac{1}{3}\)
• B. \(\frac{3}{1}\)
• C. \(\frac{1}{16}\)
• D. \(\frac{16}{1}\)

Answer 1

The Correct Option is B

To find the ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture, we will use the ideal gas law and the given data.

The ideal gas law is given by:

PV = nRT

where:

• P is the pressure in pascals (Pa)
• V is the volume in cubic meters (m^3)
• n is the number of moles
• R is the gas constant (8.3 \, JK^{-1}mol^{-1})
• T is the temperature in Kelvin (K)

Given data:

• Volume, V = 2000 \, cm^3 = 2 \times 10^{-3} \, m^3
• Temperature, T = 300 \, K
• Pressure, P = 100 \, kPa = 100 \times 10^3 \, Pa
• Total mass of gas mixture, m = 0.76 \, g

Let n_H be the number of moles of hydrogen and n_O be the number of moles of oxygen.

The mass of the gas mixture is given by:

m = n_H \cdot M_H + n_O \cdot M_O

where M_H = 2 \, g/mol (molar mass of H₂) and M_O = 32 \, g/mol (molar mass of O₂).

Thus,

0.76 = n_H \cdot 2 + n_O \cdot 32  (Equation 1)

Using the ideal gas equation for the entire gas mixture:

(n_H + n_O)R \cdot T = PV

Substituting the known values:

(n_H + n_O) \cdot 8.3 \cdot 300 = 100 \times 10^3 \cdot 2 \times 10^{-3}

(n_H + n_O) \cdot 2490 = 200

n_H + n_O = \frac{200}{2490} \approx 0.0803  (Equation 2)

Solving the two simultaneous equations (Equation 1 and Equation 2):

0.76 = n_H \cdot 2 + n_O \cdot 32

0.0803 = n_H + n_O

Let us solve these equations:

1. From Equation 2: n_O = 0.0803 - n_H
2. Substitute n_O from step 1 into Equation 1:

0.76 = n_H \cdot 2 + (0.0803 - n_H) \cdot 32

0.76 = 2n_H + 2.5696 - 32n_H

0.76 = 2.5696 - 30n_H

30n_H = 2.5696 - 0.76

30n_H = 1.8096

n_H = \frac{1.8096}{30} \approx 0.0603

n_O = 0.0803 - 0.0603 = 0.02

The ratio of the number of moles of hydrogen to oxygen is:

\frac{n_H}{n_O} = \frac{0.0603}{0.02} = \frac{3}{1}

Therefore, the ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is \frac{3}{1}.