Question

The density of a gas is 1.78 g/L at STP. The weight of one mole of gas is:
A. 39.9 g
B. 22.4 g
C. 3.56 g
D. 29 g

Answer 1

Hint: The way to solve this question is simply applying the ideal gas equation. The equation is PV=nRT. 1 mole of gas is the molar mass of a gas. Use the standard temperature and pressure values to find the required answer.

Complete step by step answer:
Let us discuss the standard temperature and pressure conditions. The standard temperature and pressure conditions or (STP) means Temperature is $0{}^\circ \text{C}$ which is 273 K. The pressure value is 1 atmosphere. The values of volume and moles are variable in it. The ideal gas equation has pressure, temperature, moles, and volume. The equation is $\text{PV=nRT}$ where R is a universal constant with value 0.0821 $\text{atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$.

One mole of gas means the molar mass of the gas. As, the substance with its molar mass or weight is considered as one mole of that substance. Like, one mole of ${{\text{H}}_{2}}$ means 2 grams of hydrogen gas, which is the molecular mass of ${{\text{H}}_{2}}$.

Let us mould the formula of the ideal gas equation in terms of density. Density is mass per unit volume. Density=$\dfrac{\text{mass}}{\text{volume}}$.

The ideal gas equation $\text{PV=nRT}$ includes $\text{n}$ which means moles. And, a mole is defined as given weight per molar mass. So, $\text{n}$ can be written as moles=$\dfrac{\text{given weight}}{\text{molar mass}}$. Now, make some rearrangements in formula, the volume which is multiplied in left-hand side, move it to right-hand side under the given weight term; $\text{P=}\dfrac{\text{given weight}}{\text{volume}}\times \dfrac{\text{RT}}{\text{molar mass}}$. The term $\dfrac{\text{given weight}}{\text{volume}}$ can be termed as density as per its formula. The formula is now moulded to $\text{P=}\dfrac{\rho \text{RT}}{\text{M}}$ where M is the molar mass of the gas.
Now, substitute the values given in the question in the particular term respectively,
$\rho $=1.78 g/L (density), R=0.0821$\text{atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$. T=273 Kelvin and P= 1 atmosphere

Apply the formula; $1=\dfrac{1.78\times 0.0821\times 273}{\text{M}}$ now, making up the further calculations, molecular mass is obtained as 39.89 grams.
- The weight of one mole of a gas is 39.98 grams.
Hence, the correct answer is option A.

Note:
There is no need to mug up all the unwanted formulae, till the time the formulae can be easily derived. In ideal gas related questions, the formula that should be learnt is $\text{PV=nRT}$. After this formulae all the other formulas can be derived.

The units need to be taken care of while solving such types of questions. There are different values of R like
0.0821$\text{atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$,
$8.314\text{ J}\text{.mo}{{\text{l}}^{-1}}\text{.}{{\text{K}}^{-1}}$ and
$1.987\text{ cal}\text{.mo}{{\text{l}}^{-1}}\text{.}{{\text{K}}^{-1}}$.
The units of R=0.0821$\text{atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$ is matching with the values of the terms in the reaction, that’s why it is used.