Question

Equal volumes of oxygen and gas R took 20 seconds and 30 seconds respectively to diffuse through the same hole under the same conditions . Find the relative molecular mass of R

Answer 1

To find the relative molecular mass of gas R using Graham's law of diffusion, we can use the following relationship:

\[
\frac{t_1}{t_2} = \sqrt{\frac{M_2}{M_1}}
\]

where:
- \( t_1 \) and \( t_2 \) are the times taken for the two gases to diffuse,
- \( M_1 \) and \( M_2 \) are the molar masses of the gases (in this case, oxygen and gas R, respectively).

Let:
- \( t_1 = 20 \) seconds (for oxygen),
- \( t_2 = 30 \) seconds (for gas R),
- \( M_1 = 32 \) g/mol (molar mass of oxygen, \( O_2 \)),

We need to find \( M_2 \) (the molar mass of gas R).

Plugging the values into the equation gives us:

\[
\frac{20}{30} = \sqrt{\frac{M_2}{32}}
\]

Simplifying the fraction on the left side:

\[
\frac{2}{3} = \sqrt{\frac{M_2}{32}}
\]

Now, squaring both sides to eliminate the square root gives:

\[
\left(\frac{2}{3}\right)^2 = \frac{M_2}{32}
\]

Calculating the left side:

\[
\frac{4}{9} = \frac{M_2}{32}
\]

Now, to find \( M_2 \), we cross-multiply:

\[
4 \cdot 32 = 9 \cdot M_2
\]

\[
128 = 9 M_2
\]

Now, divide both sides by 9:

\[
M_2 = \frac{128}{9} \approx 14.22 \text{ g/mol}
\]

Thus, the relative molecular mass of gas R is approximately **14.22 g/mol**.