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Ideal Gas Equation from KTG

A sample of p...

Question

A sample of pure $C u O$ was reduced with $H_{2}$ gas and $H_{2} O$ formed was collected in a $44.8$ L flask containing dry $N_{2}$ . At $27^{o} C$ , the total pressure containing $N_{2}$ and $H_{2} O$ was $1.0$ atm. The relative humidity in the flask was $80 %$ . The vapour pressure of water at $27^{o} C$ is $25$ mm. If $x$ grams of $C u O$ was reduced, then value of $10 x$ is :

38

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35

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20

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none of the above

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Solution

The correct option is B $38$
The reaction is as follows:
$C u O + H_{2} \to C u + H_{2} O (g)$
Volume of dry $N_{2} = 44.8$ L at $27^{o} C$
Total pressure $(N_{2} + H_{2} O) = 1.0$ atm
Vapour pressure of $H_{2} O = 25$ mm
Humidity $= 80 %$
Pressure due to $H_{2} O$ vapour $= 0.8 \times \frac{25}{760}$ mm

Moles of $H_{2} O$ vapour $= \frac{P V}{R T} = \frac{0.8 \times \frac{25}{760} \times 44.8}{0.082 \times 300}$ $= 0.0478$ mol

According to the above equation,
$1$ mol of $H_{2} O$ is obtained from $1$ mol of $C u O$ .
Therefore, moles of $C u O$ reduced $= 0.0478$ mol.
Weight of $C u O = 0.0478 \times (63.5 + 16) = 3.8$ g.

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