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Standard XII

Chemistry

Ideal Gas Law

When 2 g of...

Question

When $2$ g of gas A is introduced into an evacuated flask kept at $25^{o} C$ , the pressure was found to be $1$ atmosphere. If $3$ g of another gas B is then added to the same flask, the pressure becomes $1.5$ atm. Assuming ideal behaviour, the ratio of molecular weights $(M_{B} : M_{A})$ is :

1:3

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3:1

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2:3

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3:2

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Solution

The correct option is C 3:1
$M o l e s \propto P r e s s u r e$

For A, $\frac{2}{M w_{A}} \propto 1 a t m$ ..........(i)

Pressure of $B = 1.5 - 1 = 0.5 a t m$

For B, $\frac{3}{M w_{B}} \propto 0.5 a t m$ ........(ii)

$\frac{3}{M w_{B}} \times \frac{M w_{A}}{2} = \frac{0.5}{1}$

$\frac{M w_{A}}{M w_{B}} = 0.5 \times \frac{2}{3} = \frac{1}{3}$

$∴ M w_{B} : M w_{A} = 3 : 1$

Therefore, the option is B.

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When 2 g of a gas is introduced into an evacuated flask kept at $25^{\circ} C$ , the pressure is found to be 1 atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 0.5 atmosphere. Assuming ideal gas behaviour, calculate the ratio of molecular mass $M_{A} : M_{B}$ .

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When 2 g of gas A is introduced into an evacuated flask kept at 25oC, the pressure was found to be 1 atmosphere. If 3 g of another gas B is then added to the same flask, the pressure becomes 1.5 atm. Assuming ideal behaviour, the ratio of molecular weights (MB:MA) is :

The correct option is C 3:1Moles∝PressureFor A, 2MwA∝1 atm ..........(i)Pressure of B=1.5−1=0.5atmFor B, 3MwB∝0.5 atm ........(ii)3MwB×MwA2=0.51MwAMwB=0.5×23=13∴MwB:MwA=3:1Therefore, the option is B.

When $2$ g of gas A is introduced into an evacuated flask kept at $25^{o} C$ , the pressure was found to be $1$ atmosphere. If $3$ g of another gas B is then added to the same flask, the pressure becomes $1.5$ atm. Assuming ideal behaviour, the ratio of molecular weights $(M_{B} : M_{A})$ is :