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  • A box contains 0.90 g of liquid water in equilibrium with water vapour at <img alt="27^{\circ} \mathrm{C} \text {. }" src="/latex-image/?27%5E%7B%5Ccirc%7D%20%5Cmathrm%7BC%7D%20%5Ctext%20%7B.%

A box contains 0.90 g of liquid water in equilibrium with water vapour at 27^{\circ} \mathrm{C} \text {. } The equilibrium vapour pressure of water at 27^{\circ} \mathrm{C} is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be ________liter. [nearest integer]

\text { (Given: } \mathrm{R}=0.082 \mathrm{~L} \operatorname{atm~} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \text { ) }

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.) 

Option: 1

29


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given, all of the water vapor evaporates upon increasing the volume.

So, amount of water vapour formed will be \mathrm{0.9 ~g} and the pressure in the container will be \mathrm{32 ~Torr}.

Therefore, from ideal gas equation, we have

\mathrm{P V=n R T}

\mathrm{V=\frac{n R T}{P}=\frac{0.90 \times 0.082 \times 300 \times 760}{18 \times 32}=29.21 \mathrm{~L}}

                    \mathrm{\simeq 29 L} ( nearest integer )

Hence answer is 29

Posted by

Rishi

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