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(A) According to the Ideal Gas Law,$PV = nRT$. Since the volume $V$ and temperature $T$ are constant,the pressure $P$ is directly proportional to the

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$1 : 3$

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(A) According to the Ideal Gas Law,$PV = nRT$. Since the volume $V$ and temperature $T$ are constant,the pressure $P$ is directly proportional to the number of moles $n$ $(P \propto n)$.Let the molecular weights of the two gases be $M_A$ and $M_B$.For the first gas $A$,the number of moles is $n_A = \frac{2}{M_A}$. Given $P_A = 1 \, atm$,we have $\frac{2}{M_A} \propto 1 \dots (i)$.When the second gas $B$ is added,the total pressure becomes $1.5 \, atm$. The partial pressure of gas $B$ is $P_B = P_{total} - P_A = 1.5 - 1 = 0.5 \, atm$.For the second gas $B$,the number of moles is $n_B = \frac{3}{M_B}$. Thus,$\frac{3}{M_B} \propto 0.5 \dots (ii)$.Dividing equation $(ii)$ by equation $(i)$:$\frac{3/M_B}{2/M_A} = \frac{0.5}{1}$$\frac{3}{M_B} 	imes \frac{M_A}{2} = 0.5$$\frac{M_A}{M_B} = 0.5 	imes \frac{2}{3} = \frac{1}{3}$Therefore,the ratio of the molecular weights $M_A : M_B$ is $1 : 3$.

When $2 \, g$ of a gas are introduced into an evacuated flask kept at $25 \, ^\circ C$,the pressure is found to be $1 \, atm$. If $3 \, g$ of another gas is added to the same flask,the pressure becomes $1.5 \, atm$. The ratio of the molecular weights of these gases will be:

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