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At  0^{\circ} \mathrm{C} , the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide?

Option: 1

60 \mathrm{~g} / \mathrm{mol}


Option: 2

70 \mathrm{~g} / \mathrm{mol}


Option: 3

75 \mathrm{~g} / \mathrm{mol}


Option: 4

65 \mathrm{~g} / \mathrm{mol}


Answers (1)

best_answer

Let, density of nitrogen and gaseous oxide be  \mathrm{\rho_1~ and ~\rho_2} 

Using formula,     \mathrm{\rho=\frac{P M}{R T}}

Pressure of nitrogen,
\mathrm{ P=5 \text { bars }=5 \times 0.987 \mathrm{~atm} }       

molar mass of nitrogen,
M=28 

\mathrm{ \left.\therefore \rho_1 \text { (density of nitrogen }\right)=\frac{5 \times 0.987 \times 28}{R \times 273} \\ }

\mathrm{ \therefore \rho_2 \text { (density of gaseous oxide) }=\frac{2 \times 0.987 \times x}{R \times 273} }

As per question, both densities are equal at  \mathrm{0^{\circ} \mathrm{C} }

\mathrm{\therefore \frac{2 \times 0.987 \times x}{R \times 273}=\frac{5 \times 0.987 \times 28}{R \times 273} \\ }
\mathrm{ \Rightarrow x=\frac{5 \times 28}{2}=70 \mathrm{~g} / \mathrm{mol}(\text { option b) } }
 

Posted by

Deependra Verma

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