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Standard XII

Chemistry

Ideal Gas Equation

The piston sh...

Question

The piston shown in the figure is moved downwards such that the vapour volume is decreased by 246.3cc. How many grams of $H_{2} O$ is condensed? (Assume aqueous tension at 27 $^{\circ}$ C is 360 mm Hg)

1.8 gm

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0.09 gm

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0.9 gm

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0.18 gm

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Solution

The correct option is B
0.09 gm

PV = $\frac{W}{M} R T \Rightarrow W_{H_{2} O} = \frac{P V M}{R T}$

$W_{H_{2} O} = \frac{360}{720} \times \frac{246.3}{1000} \times \frac{18}{0.0821 \times 300} \Rightarrow 0.09 g m .$

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Similar questions

The piston shown in the figure is moved downwards such that the vapour volume is decreased by 246.3cc. How many grams of $H_{2} O$ is condensed? (Assume aqueous tension at 27 $^{\circ}$ C is 360 mm Hg)

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Q. 0.157 g of a certain gas collected over water occupied a volume of 135 ml at

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The piston shown in the figure is moved downwards such that the vapour volume is decreased by 246.3cc. How many grams of H2O is condensed? (Assume aqueous tension at 27∘C is 360 mm Hg)

The correct option is B 0.09 gm PV = WMRT⇒WH2O=PVMRT WH2O=360720×246.31000×180.0821×300⇒0.09 gm.

The piston shown in the figure is moved downwards such that the vapour volume is decreased by 246.3cc. How many grams of $H_{2} O$ is condensed? (Assume aqueous tension at 27 $^{\circ}$ C is 360 mm Hg)

The correct option is B
0.09 gm

PV = $\frac{W}{M} R T \Rightarrow W_{H_{2} O} = \frac{P V M}{R T}$

$W_{H_{2} O} = \frac{360}{720} \times \frac{246.3}{1000} \times \frac{18}{0.0821 \times 300} \Rightarrow 0.09 g m .$