Question

A gas jar of 10 litre volume filled with O, at 300 K is connected
to glycerine manometer. The manometer shown 5 m difference
in the level as shown in figure. What will be the number of
moles of O2 in the gas jar?
(Given decenine = 2.72 g/mL; dacsony = 13.6 g/mL)
Gas
5 m
102
3ook
(A) 0.64 mol
(B) 0.4 mol
(C) 0.94 mol
(D) 0.36 mol

Answer 1

Answer : The correct option is, (C) 0.94 mol

Explanation : Given,

Volume of container = 10 L

Temperature = 300 K

Density of glycerine = 2.72 g/mL

Density of mercury = 13.6 g/mL

Height difference by glycerine = 5 m

First we have to calculate the height of mercury.

Formula used : $P=h\rho g$

where,

P is pressure, h is height, g is acceleration due to gravity and $\rho$ is density.

The expression for gylcerine and mercury will be:

$P_{(Hg)}=h_{(Hg)}\rho_{(Hg)}g\\\\P_{(gly)}=h_{(gly)}\rho_{(gly)}g$

So,

$h_{(Hg)}\rho_{(Hg)}g=h_{(gly)}\rho_{(gly)}g$

$h_{(Hg)}\rho_{(Hg)}=h_{(gly)}\rho_{(gly)}$

Now put all the given values in this expression, we get:

$h_{Hg}\times (13.6g/mL)=(5m)\times (2.72g/mL)$

$h_{Hg}=1m$

Now we have to calculate the pressure of the gas.

$P_{gas}=0.76dg+1dg=1.76dg=1760mmHg=\frac{1760}{760}atm$

conversion used : (1 atm = 760 mmHg)

Now we have to calculate the moles of gas.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = $\frac{1760}{760}atm$

V = Volume of gas = 10 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 300 K

Putting values in above equation, we get:

$(\frac{1760}{760}atm)\times 10L=n\times (0.0821L.atm/mol.K)\times 300K\\\\n=0.94mole$

Therefore, the moles of $O_2$ in the gas jar will be 0.94 mol