The density of a gas A is three times that of a gas B. It the molecular mass of A is M, the molecular mass of B is

Options

A. 3M
B. `sqrt(3M)`
C. `(M)/(3)`
D. `(M)/(sqrt(3))`

Explanation

To solve the problem, we will use the relationship between the density of gases and their molecular masses. Here's a step-by-step solution:

### Step 1: Understand the relationship between density and molecular mass
The density of a gas is related to its molecular mass and the conditions of temperature and pressure. The formula for density (D) can be expressed as:
\[ D = \frac{m}{V} \]
where \( m \) is the mass of the gas and \( V \) is the volume.

### Step 2: Use the ideal gas law
The ideal gas law states:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
where \( M \) is the molecular mass of the gas.

### Step 3: Substitute the number of moles into the ideal gas law
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]
Rearranging this gives:
\[ P = \frac{mRT}{MV} \]

### Step 4: Relate pressure to density
We can express density in terms of mass and volume:
\[ D = \frac{m}{V} \]
Thus, we can rewrite the pressure equation as:
\[ P = D \cdot \frac{RT}{M} \]

### Step 5: Set up the equations for gases A and B
Let the density of gas A be \( D_A \) and the density of gas B be \( D_B \). According to the problem, we have:
\[ D_A = 3D_B \]
Let the molecular mass of gas A be \( M_A = M \) and the molecular mass of gas B be \( M_B \).

Using the pressure equation for both gases:
\[ P = D_A \cdot \frac{RT}{M_A} \]
\[ P = D_B \cdot \frac{RT}{M_B} \]

### Step 6: Equate the two expressions for pressure
Since both gases are at the same temperature and pressure, we can set the two equations equal to each other:
\[ D_A \cdot \frac{RT}{M_A} = D_B \cdot \frac{RT}{M_B} \]

### Step 7: Cancel common terms
Cancel \( RT \) from both sides:
\[ D_A \cdot \frac{1}{M_A} = D_B \cdot \frac{1}{M_B} \]

### Step 8: Substitute the relationship of densities
Substituting \( D_A = 3D_B \):
\[ 3D_B \cdot \frac{1}{M_A} = D_B \cdot \frac{1}{M_B} \]

### Step 9: Simplify the equation
Cancel \( D_B \) (assuming \( D_B \neq 0 \)):
\[ 3 \cdot \frac{1}{M_A} = \frac{1}{M_B} \]

### Step 10: Solve for \( M_B \)
Rearranging gives:
\[ M_B = \frac{M_A}{3} \]
Since \( M_A = M \):
\[ M_B = \frac{M}{3} \]

Thus, the molecular mass of gas B is \( \frac{M}{3} \).

### Final Answer:
The molecular mass of gas B is \( \frac{M}{3} \).

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