Question

2 g of a gas A introduced into an evacuated flask at 25 ⁰C. The pressure of the gas is 1 atm. Now 3 g of another gas B is introduced in the same flask so total pressure becomes 1.5 atm. The ratio of molecular mass of A​ and B is:
$
  A:\dfrac{3}{1} \\
  B:\dfrac{1}{3} \\
  C:\dfrac{1}{4} \\
  D:\dfrac{2}{3} \\
$

Answer 1

Hint: The partial pressure of a gas refers to the pressure which is exerted by a gas in the volume being occupied by a mixture of gases. The summation of the partial pressure of all the gases in a mixture is equal to the total pressure.

Complete Step by step answer: It is given that when 2 g of a gas A has been introduced into an evacuated flask at 25 ⁰C , the pressure of the gas is 1 atm. That means,
Partial pressure of the gas A = 1 atm.
We know that:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}$
Using the above formula we will calculate the number of moles of gas A, nA.
Mass of gas A = 2 g (Given)
Molar mass of gas A = MA (Assumption)
$\therefore {N_A} = \dfrac{2}{{{M_A}}}$
Now it is given that 3 g of another gas B has been added into the same flask and total pressure becomes 1.5 atm. This means:
Mass of gas B = 3 g (Given)
Molecular mass of gas B = MB (Assumption)
Using the aforementioned formula, we will now calculate the number of moles of gas B, nB:
${N_B} = \dfrac{3}{{{M_B}}}$
\[\begin{array}{*{20}{l}}
  {Total{\text{ }}pressure = Partial{\text{ }}pressure{\text{ }}of{\text{ }}gas{\text{ }}A + Partial{\text{ }}pressure{\text{ }}of{\text{ }}gas{\text{ }}B} \\
  {Partial{\text{ }}pressure{\text{ }}of{\text{ }}gas{\text{ }}B = Total{\text{ }}pressure-Partial{\text{ }}pressure{\text{ }}of{\text{ }}gas{\text{ }}A} \\
  {Partial{\text{ }}pressure{\text{ }}of{\text{ }}gas{\text{ }}B = 1.5 - 1 = 0.5{\text{ }}atm}
\end{array}\]
At constant temperature and volume:
$\dfrac{{{P_A}}}{{{n_A}}} = \dfrac{{{P_B}}}{{{n_B}}}{\text{ }}(from{\text{ }}PV = nRT)$
$\therefore \dfrac{1}{{\dfrac{2}{{{M_A}}}}} = \dfrac{{0.5}}{{\dfrac{3}{{{M_B}}}}}$
\[
  {M_A} = \dfrac{{0.5{M_B}}}{3} \times 2 \\
  \Rightarrow \dfrac{{{M_A}}}{1} = \dfrac{{{M_B}}}{3} \\
  \Rightarrow \dfrac{{{M_A}}}{{{M_B}}} = \dfrac{1}{3} \\
\]
Thus, the ratio of molecular mass of A​ and B is 1:3.

Hence, the correct answer is Option B.

Note: Dalton's law states that the total pressure is the sum of individual pressures at a constant temperature. This empirical law was invented by John Dalton in 1801. This law is basically correlated to the ideal gas law.