Question

The density of a gas at STP is 2.5 g/L. the weight of $6.02 \times {10^{23}}$ molecules of the gas is:
A.28 g
B.14 g
C.42 g
D.56 g

Answer 1

We know that 1 mole of any ideal gas occupies 22.4 L at STP conditions and its given that the density is 2.5 g/L.

Complete step by step answer:
Given,
Density of a gas at STP = 2.5 g/L
$6.02 \times {10^{23}}$ molecules contain 1 mole of gas = molar mass of the gas
We know, 1 mole of any ideal gas occupies 22.4 L at STP conditions (mole volume)
Therefore, 1 litre of gas has a mass 2.5 g at STP
∴ 1 mole of the gas will have the mass of $= 22.4L \times \dfrac{{2.5g}}{{1L}}$ = 56 g
So, if one mole of this gas has a mass of 56 g, we can say that its molar mass is equal to 56 g.
So, if the density of a gas at STP is 2.5 g/L. the weight of $6.02 \times {10^{23}}$ molecules of the gas is 56 g.

Therefore, the correct answer is option (D)

Note: The alternate method to solve this problem is,
We know,
$6.02 \times {10^{23}}$ molecules contain 1 mole of gas = molar mass of the gas
$Density = \dfrac{{Mass}}{{Volume}}$
$\Rightarrow$ $2.5 = \dfrac{{Molar\,mass}}{{22.4}}$
$\Rightarrow$ Molar mass $= 2.5 \times 22.4$ = 56 g