A gas jar of $10$ litre volume filled with $O_{2}$ at $300 K$ is connected to glycerine manometer. The manometer shows $5 m$ difference in the level as shown in figure. What will be the number of moles of $O_{2}$ in the gas jar?
(Give $d_{g l y c e r i n e} = 2.72 g / m L; d_{m e r c u r y} = 13.6 g / m L)$

0.64 m o l

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0.4 m o l

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0.94 m o l

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0.36 m o l

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Solution

The correct option is D $0.94 m o l$
$(h \times d \times g)_{H g} = (h \times d \times g)_{g l y c e r i n e}$
$(h \times 13.6)_{H g} = (5 \times 2.72)_{g l y c e r i n e}$
$h_{H g} = 1 m$
$P_{g a s} = (1 + 0.76) m$
$= 1760 m m H g$
$P V = n R T$
$\frac{1760}{760} \times 10 = n \times 0.0821 \times 300$
$n = 0.94 m o l$

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Similar questions

Q. Calculate the number of moles of gas present in the container of volume

10 l i t

300 K

. If the manometer containing glycerine shows

5 m

difference in level as shown in the diagram.
Given

d_{g l y c e r i n e} = 2.72 g m / m l, d_{m e r c u r y} = 13.6 g m / m l

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