Three gases A,B and C are at same temperture if their r.m.s speed are in the ratio 1:`1/sqrt2:1/sqrt3` then their molar masses will be in the ratio :

Options

A. `1:2:3`
B. `3:2:1`
C. `1:sqrt2:sqrt3`
D. `sqrt3:sqrt2:1`

Explanation

To solve the problem, we need to find the molar masses of gases A, B, and C based on their root mean square (r.m.s) speeds given in the ratio 1 : 1/√2 : 1/√3.

### Step-by-Step Solution:

1. **Understand the formula for r.m.s speed**:
The r.m.s speed (V_rms) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

2. **Set up the r.m.s speed ratios**:
For gases A, B, and C, we denote their r.m.s speeds as:
\[ V_A : V_B : V_C = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \]

3. **Express the ratios in terms of molar masses**:
Since all gases are at the same temperature, we can express their r.m.s speeds in terms of their molar masses:
\[ \frac{V_A}{V_B} = \sqrt{\frac{M_B}{M_A}}, \quad \frac{V_A}{V_C} = \sqrt{\frac{M_C}{M_A}}, \quad \frac{V_B}{V_C} = \sqrt{\frac{M_C}{M_B}} \]

4. **Square the ratios**:
Squaring the ratios gives us:
\[ \left(\frac{V_A}{V_B}\right)^2 = \frac{M_B}{M_A}, \quad \left(\frac{V_A}{V_C}\right)^2 = \frac{M_C}{M_A}, \quad \left(\frac{V_B}{V_C}\right)^2 = \frac{M_C}{M_B} \]

5. **Substituting the given ratios**:
From the given ratio of r.m.s speeds:
\[ \frac{V_A}{V_B} = 1, \quad \frac{V_A}{V_C} = \frac{1}{\sqrt{3}}, \quad \frac{V_B}{V_C} = \frac{1}{\sqrt{2}} \]
We can equate these ratios:
\[ 1^2 = \frac{M_B}{M_A} \implies M_A = M_B \]
\[ \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{M_C}{M_A} \implies M_C = \frac{M_A}{3} \]
\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{M_C}{M_B} \implies M_C = \frac{M_B}{2} \]

6. **Finding the ratios**:
From \( M_A = M_B \) and \( M_C = \frac{M_A}{3} \):
- Let \( M_A = x \)
- Then \( M_B = x \)
- And \( M_C = \frac{x}{3} \)

The ratio of molar masses \( M_A : M_B : M_C \) becomes:
\[ x : x : \frac{x}{3} = 1 : 1 : \frac{1}{3} = 3 : 3 : 1 \]

7. **Final ratio**:
To express the ratio in a more standard form, we can multiply through by 3:
\[ M_A : M_B : M_C = 3 : 3 : 1 \]

### Conclusion:
The molar masses of gases A, B, and C are in the ratio \( 1 : 2 : 3 \).