Two samples of gases 'a' and 'b' are at the same temperature. The molecules of 'a' are travelling 4 times faster than molecules of 'b'. The ratio of `M_(a)//M_(b)` will be

Options

A. `1//4`
B. `16//1`
C. `4//1`
D. `1//16`

Answer

To solve the problem, we will use the relationship between the speed of gas molecules and their molecular masses, as described by Graham's Law of Effusion/Diffusion.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Let the speed of gas A be \( r_A \) and the speed of gas B be \( r_B \).
- According to the problem, \( r_A = 4r_B \). This means that the molecules of gas A are traveling 4 times faster than those of gas B.

2. **Apply Graham's Law**:
- Graham's Law states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass (molecular mass).
- Mathematically, this can be expressed as:
\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]
- Here, \( M_A \) is the molar mass of gas A and \( M_B \) is the molar mass of gas B.

3. **Substitute the Known Values**:
- From the problem, we know \( \frac{r_A}{r_B} = 4 \).
- Therefore, we can substitute this into Graham's Law:
\[ 4 = \sqrt{\frac{M_B}{M_A}} \]

4. **Square Both Sides**:
- To eliminate the square root, we square both sides of the equation:
\[ 4^2 = \frac{M_B}{M_A} \]
- This simplifies to:
\[ 16 = \frac{M_B}{M_A} \]

5. **Find the Ratio of Molecular Masses**:
- Rearranging the equation gives us:
\[ \frac{M_A}{M_B} = \frac{1}{16} \]
- Thus, the ratio of the molecular masses of gases A and B is:
\[ \frac{M_A}{M_B} = 1:16 \]

### Final Answer:
The ratio of \( \frac{M_A}{M_B} \) is \( 1:16 \).

---