In what ratio by mass, sulphur trioxide and nitrogen should be mixed so that partial pressure exerted by each gas is same ?

Options

A. `7: 20`
B. `7:40`
C. `20 :7`
D. `40:7`

Explanation

To solve the problem of finding the mass ratio of sulfur trioxide (SO3) and nitrogen (N2) that should be mixed so that the partial pressure exerted by each gas is the same, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Partial Pressure**: The partial pressure of a gas in a mixture can be expressed in terms of its mole fraction and the total pressure. For two gases, the condition for equal partial pressures can be set up using their mole fractions.

2. **Define Mole Fraction**: The mole fraction (X) of a gas is defined as:
\[ X_{SO3} = \frac{n_{SO3}}{n_{SO3} + n_{N2}} \]
\[ X_{N2} = \frac{n_{N2}}{n_{SO3} + n_{N2}} \]
where \(n\) represents the number of moles of each gas.

3. **Set Up the Equation for Equal Partial Pressures**: Since we want the partial pressures to be equal, we can set the mole fractions equal to each other:
\[ X_{SO3} = X_{N2} \]

4. **Express Moles in Terms of Mass**: The number of moles can be expressed in terms of mass (W) and molar mass (M):
\[ n_{SO3} = \frac{W_1}{M_{SO3}} \quad \text{and} \quad n_{N2} = \frac{W_2}{M_{N2}} \]
where \(W_1\) and \(W_2\) are the masses of SO3 and N2, and \(M_{SO3}\) and \(M_{N2}\) are their molar masses.

5. **Substitute Molar Masses**: The molar mass of SO3 is 80 g/mol (32 for sulfur and 48 for three oxygens) and the molar mass of N2 is 28 g/mol. Substitute these values into the mole fraction equations:
\[ X_{SO3} = \frac{W_1/80}{(W_1/80) + (W_2/28)} \]
\[ X_{N2} = \frac{W_2/28}{(W_1/80) + (W_2/28)} \]

6. **Set the Mole Fractions Equal**: Since \(X_{SO3} = X_{N2}\), we can set the two expressions equal to each other:
\[ \frac{W_1/80}{(W_1/80) + (W_2/28)} = \frac{W_2/28}{(W_1/80) + (W_2/28)} \]

7. **Cross Multiply**: Cross multiplying gives:
\[ W_1 \cdot (W_2/28) = W_2 \cdot (W_1/80) \]

8. **Rearranging the Equation**: Rearranging the equation leads to:
\[ 80W_1 = 28W_2 \]

9. **Finding the Mass Ratio**: To find the ratio \(W_1 : W_2\), we can express it as:
\[ \frac{W_1}{W_2} = \frac{28}{80} = \frac{7}{20} \]

### Final Answer:
The mass ratio of sulfur trioxide to nitrogen should be \(7 : 20\).