At S.T.P., the volume of 7.5 g of a gas is 5 | vedclass

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(A) Given, mass of gas $(W) = 7.5 \ g$. Volume of gas at $STP$ $(V) = 5.6 \ L$. Since, moles of gas at $STP = \frac{V(L)}{22.4 \ L} = \frac{W}{M}$, wh

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$NO$

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(A) Given, mass of gas $(W) = 7.5 \ g$. Volume of gas at $STP$ $(V) = 5.6 \ L$. Since, moles of gas at $STP = \frac{V(L)}{22.4 \ L} = \frac{W}{M}$, where $M$ is the molar mass of the gas. Therefore, $M = \frac{W 	imes 22.4}{V}$. $M = \frac{7.5 	imes 22.4}{5.6} = 30.00 \ g \ mol^{-1}$. Among the given options, the molar mass $(M)$ of: $(A) \ NO = 14 + 16 = 30.00 \ g \ mol^{-1}$. $(B) \ N_{2}O = 28 + 16 = 44.00 \ g \ mol^{-1}$. $(C) \ CO = 12 + 16 = 28.00 \ g \ mol^{-1}$. $(D) \ CO_{2} = 12 + 32 = 44.00 \ g \ mol^{-1}$. Since the molar mass of $NO$ is $30 \ g \ mol^{-1}$, the given gas is $NO$. Hence, $A$ is the correct option.

At $S.T.P.$, the volume of $7.5 \ g$ of a gas is $5.6 \ L$. The gas is

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