Find number of moles of gas $A$ , present in the container of volume $10 L$ and temperature $300 K$ . Use shown below. [ $d_{g l y c e r i n} = 2.72 g / m l, d_{m e r c u r y} = 13.6 g / m l$ ]

0.94 m o l e

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0.49 m o l e

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0.64 m o l e

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0.66 m o l e

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Solution

The correct option is B $0.94 m o l e$
$Δ P = h ρ g = 5 \times 2.72 \times 9.8 = 133.28 k P a$
Convert the unit in atm
$Δ P = \frac{133.28}{101.3} = 1.315 a t m$
$P = 1 a t m + 1.315 a t m = 2.315 a t m$
$n = \frac{P V}{R T} = \frac{2.315 \times 10}{0.0821 \times 300} = 0.94 m o l$

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Similar questions

Calculate the number of moles of gas present in the container of volume $10 L$ at $300 K$ . If the manometer containing glycerin shows $5 m$ difference in level as shown in the diagram. calculate your answers up to two decimal points.

(Given: $d_{glycerin} = 2.72 g/mL; d_{mercury} = 13.6 g/mL Tale R = 0.08 L a t m m o l^{- 1} K^{- 1}$ )