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

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