Question

If 30 g of $N{O_x}\left( g \right)$ occupies 22.4 L at S.T.P then the value of x is
A) 2
B) 1
C) 0.5
D) 2.5

Answer 1

Hint:$N{O_x}$has a mass of 30 g and a molar volume of 22.4 L at S.T.P which means 30 g is its molar mass as it occupies same volume as the volume occupied by any gas at S.T.P. Then, we’ll calculate the molar mass of $N{O_x}$ by multiplying the subscript of N with its molar mass and subscript of O with its molar mass and then equate it with 30 g.

Complete step by step answer:
Standard temperature and pressure (S.T.P) are standard sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data. The most commonly used standards are International Union of Pure and Applied Chemistry (IUPAC). The volume occupied by 1 gram molecule or 1 mole of a dry gas at S.T.P is called gram molecular volume or molar volume. The value of 1 gram molecular volume at S.T.P is 22.4 L or 22.4 $d{m^3}$ or 22400 $c{m^3}$ . It is given that 30 g of $N{O_x}$ occupies 22.4 L at S.T.P and $N{O_x}$ is present in the gaseous state. It means that 30 g is the molar mass of $N{O_x}$ as it occupies a volume of 22.4 L. Now, we’ll calculate molar mass of $N{O_x}$.
Molar mass of $N{O_x}$ $\left( m \right)$ = $1 \times N + x \times O$
$m = 1 \times 14 + x \times 16$ [molar mass of N (nitrogen) is 14 g and O (oxygen) is 16 g]
$m = 14 + 16x$
$30 = 14 + 16x$
$30 - 14 = 16x$
$16 = 16x$
$x = \dfrac{{16}}{{16}}$
$x = 1$
Therefore, option B is correct.

Note: Remember that at S.T.P i.e., at standard temperature and pressure any dry gas occupies a volume of 22.4 L which means at that volume the mass of the gas would be considered as its molar mass. Also, memorize the molar mass of nitrogen and oxygen.