If the density of a gas A is 1.5 times that of B, then the molecular mass of A is M. The molecular mass of B will be

Options

A. 1.5 M
B. `M//1.5`
C. 3 M
D. `M//3`

Explanation

To solve the problem step by step, we will use the relationship between the density of gases and their molecular masses.

### Step-by-Step Solution:

1. **Understand the Relationship**:
We know that the density (d) of a gas is related to its molecular mass (M) and the conditions of temperature (T) and pressure (P) through the ideal gas equation. The formula can be rearranged to express molecular mass in terms of density:
\[ M = \frac{dRT}{P} \]
where R is the universal gas constant.

2. **Set Up the Equations**:
For gas A, we can write:
\[ M_A = \frac{d_A RT}{P} \]
For gas B, we can write:
\[ M_B = \frac{d_B RT}{P} \]

3. **Relate the Densities**:
According to the problem, the density of gas A is 1.5 times that of gas B:
\[ d_A = 1.5 d_B \]

4. **Substitute the Density Relation**:
Substitute \(d_A\) into the equation for \(M_A\):
\[ M_A = \frac{(1.5 d_B) RT}{P} \]

5. **Express Molecular Mass of A in terms of B**:
Since we know \(M_A = M\), we can set up the equation:
\[ M = \frac{(1.5 d_B) RT}{P} \]

6. **Express Molecular Mass of B**:
From the equation for \(M_B\):
\[ M_B = \frac{d_B RT}{P} \]

7. **Relate \(M\) and \(M_B\)**:
Now, we can relate \(M\) to \(M_B\):
\[ M_B = \frac{d_B RT}{P} = \frac{M}{1.5} \]

8. **Final Expression for Molecular Mass of B**:
Therefore, the molecular mass of gas B can be expressed as:
\[ M_B = \frac{M}{1.5} \]

### Conclusion:
The molecular mass of gas B is \(\frac{M}{1.5}\).

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