Certainly! Let's solve this problem step by step.
Step 1: Understanding Graham's Law of Effusion
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be written as:
[tex]\[
\frac{\text{Rate}_P}{\text{Rate}_Q} = \sqrt{\frac{\text{Molar mass}_Q}{\text{Molar mass}_P}}
\][/tex]
Step 2: Plugging in the Given Data
We are given:
- The rate of diffusion of gas [tex]\(P\)[/tex] ([tex]\(\text{Rate}_P\)[/tex]) = [tex]\(12 \, \text{cm}^3/\text{s}\)[/tex]
- The rate of diffusion of gas [tex]\(Q\)[/tex] ([tex]\(\text{Rate}_Q\)[/tex]) = [tex]\(7.2 \, \text{cm}^3/\text{s}\)[/tex]
- The molar mass of gas [tex]\(P\)[/tex] ([tex]\(\text{Molar mass}_P\)[/tex]) = [tex]\(16 \, \text{g/mol}\)[/tex]
Step 3: Applying Graham's Law
Using Graham's Law, we can write:
[tex]\[
\frac{12}{7.2} = \sqrt{\frac{\text{Molar mass}_Q}{16}}
\][/tex]
Step 4: Simplifying the Equation
First, calculate the ratio on the left side:
[tex]\[
\frac{12}{7.2} = \frac{5}{3}
\][/tex]
This gives us:
[tex]\[
\frac{5}{3} = \sqrt{\frac{\text{Molar mass}_Q}{16}}
\][/tex]
Step 5: Solving for [tex]\(\text{Molar mass}_Q\)[/tex]
Next, square both sides to eliminate the square root:
[tex]\[
\left(\frac{5}{3}\right)^2 = \frac{\text{Molar mass}_Q}{16}
\][/tex]
Calculate [tex]\(\left(\frac{5}{3}\right)^2\)[/tex]:
[tex]\[
\left(\frac{5}{3}\right)^2 = \frac{25}{9}
\][/tex]
So we have:
[tex]\[
\frac{25}{9} = \frac{\text{Molar mass}_Q}{16}
\][/tex]
Now, solve for [tex]\(\text{Molar mass}_Q\)[/tex] by multiplying both sides by 16:
[tex]\[
\text{Molar mass}_Q = \frac{25}{9} \times 16
\][/tex]
Calculate the value:
[tex]\[
\text{Molar mass}_Q = \frac{25 \times 16}{9} = \frac{400}{9} \approx 44.44 \, \text{g/mol}
\][/tex]
Conclusion
The molar mass of gas [tex]\(Q\)[/tex] is approximately [tex]\(44.44 \, \text{g/mol}\)[/tex].
