A gaseous mixture of 3 gases \(A, B\) and \(C\) has a pressure \(10 \mathrm{~atm}\). The total number of…
A gaseous mixture of 3 gases \(A, B\) and \(C\) has a pressure \(10 \mathrm{~atm}\). The total number of moles are 10. If partial pressure of gases \(A\) and \(B\) are \(3 \mathrm{~atm}\) and \(1 \mathrm{~atm}\) respectively and if molar mass of gas \(C\) is 2 . Calculate the weight of \(C\) in the mixture.
\(12 \mathrm{~g}\)
\(24 \mathrm{~g}\)
\(8 \mathrm{~g}\)
\(2 \mathrm{~g}\)
Solution
(i) \(p=p_A+p_B+p_C\) (Dalton's law)
\((\because p=\text { total pressure, } p^{\prime}=\text { partial }\)
pressure,
\(\begin{aligned}
& n=\text { number of moles }) \\
\Rightarrow & p_C=p-\left(p_A+p_B\right)=10-(3+1)=6 \mathrm{~atm}
\end{aligned}\)
(ii)
\(\begin{aligned}
& p_C=\frac{n_C}{n_A+n_B+n_C} \times p \\
& \Rightarrow \quad \frac{n_C}{10} \times 10=n_C \\
& \Rightarrow \quad n_C=p_C=6
\end{aligned}\)
(iii) Weight of \(C\) in the mixture \(=n_C \times\) molar mass of \(C\) \(=6 \times 2=12 \mathrm{~g}\)