The density of a certain gaseous oxide at 3-2 bar pressure and 47 -C is same as that of O2 at 27 -C and 4-2 bar pressure- The molecular mass of the gaseous oxide in g is-

Given, d_O_2 = d_oxide We know, Density, d=PMRT ∴P_1M_1RT_1 = P_2M_2RT_2 nbsp; Here, P_1 = 4.2 bar, T_1=27 ^∘C = 300 K P_2 = nbsp;3.2 bar, T_2 = nbsp;47 ^ ...

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Standard XII

Chemistry

Law of Radioactivity

The density o...

Question

The density of a certain gaseous oxide at $3.2 bar$ pressure and $47^{\circ} C$ is same as that of $O_{2}$ at $27^{\circ} C$ and $4.2 bar$ pressure. The molecular mass of the gaseous oxide (in $g$ ) is:

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Solution

Given,
$d_{O_{2}} = d_{o x i d e}$
We know,
$Density, d = \frac{P M}{R T}$
$∴ \frac{P_{1} M_{1}}{R T_{1}} = \frac{P_{2} M_{2}}{R T_{2}}$
Here, $P_{1} = 4.2 bar, T_{1} = 27^{\circ} C = 300 K$
$P_{2} = 3.2 bar, T_{2} = 47^{\circ} C = 320 K$
$∴ \frac{4.2 \times 32}{300} = \frac{3.2 \times M_{2}}{320} \Rightarrow M_{2} = 44.8 g$

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The density of a certain gaseous oxide at 3.2 bar pressure and 47 ∘C is same as that of O2 at 27 ∘C and 4.2 bar pressure. The molecular mass of the gaseous oxide (in g) is:

Given, dO2=doxide We know, Density, d=PMRT ∴P1M1RT1=P2M2RT2 Here, P1=4.2 bar, T1=27 ∘C=300 K P2=3.2 bar, T2=47 ∘C=320 K ∴ 4.2×32300=3.2×M2320⇒M2=44.8 g

The density of a certain gaseous oxide at $3.2 bar$ pressure and $47^{\circ} C$ is same as that of $O_{2}$ at $27^{\circ} C$ and $4.2 bar$ pressure. The molecular mass of the gaseous oxide (in $g$ ) is: