The mass of 4.48 dm^3 of a certain gas is 5. | vedclass

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(C) At $STP$,$1 \ mole$ of an ideal gas occupies $22.4 \ dm^3$ volume.Given volume of gas = $4.48 \ dm^3$.Number of moles $(n)$ = $\frac{	ext{Given V

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$N_2$

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(C) At $STP$,$1 \ mole$ of an ideal gas occupies $22.4 \ dm^3$ volume.Given volume of gas = $4.48 \ dm^3$.Number of moles $(n)$ = $\frac{	ext{Given Volume}}{	ext{Molar Volume at STP}} = \frac{4.48 \ dm^3}{22.4 \ dm^3/mol} = 0.2 \ mol$.We know that $n = \frac{	ext{Mass}}{	ext{Molar Mass (M)}}$.$0.2 \ mol = \frac{5.6 \ g}{M}$.$M = \frac{5.6 \ g}{0.2 \ mol} = 28 \ g/mol$.The molar mass of $N_2$ is $2 	imes 14 = 28 \ g/mol$.Therefore,the gas is $N_2$.

The mass of $4.48 \ dm^3$ of a certain gas is $5.6 \ g$ at $STP$. Assuming ideal behaviour,identify the probable gas from the following.

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