Question

Which sample at STP has the same number of molecules as 5 litre of NO2(g) at STP ? (A) 5 gram of H2(g) (B) 5 litre of CH4(g) (C) 5 moles of O2(g) (D) 5 × 1023 molecules of CO2(g)

Answer 1

the correct answer is B ) 5 litre of CH4

Answer 2

Answer:  (B) 5 litre of CH4(g)

Explanation: According to Avogadro's law, 1 mole of every gas occupies 22.4 Litres at STP and contains avogadro's number of particles and weighs equal to the molecular mass.

Thus 22.4 L of $NO_2$ contains $6.023\times 10^{23}$ molecules at STP

5 L of $NO_2$ contains=$\frac{6.023\times 10^{23}}{22.4}\times 5=1.34\times 10^{23}$ molecules at STP

a) 5 gram of $H_2$

2 g of  $H_2$ contains $6.023\times 10^{23}$ molecules at STP

5 g  $H_2$ contains= $\frac{6.023\times 10^{23}}{2}\times 5=15.05\times 10^{23}$ molecules at STP

b) 5 litre of $CH_4$

Thus 22.4 L of $CH_4$ contains $6.023\times 10^{23}$ molecules at STP

5 L of $CH_4$ contains=$\frac{6.023\times 10^{23}}{22.4}\times 5=1.34\times 10^{23}$ molecules at STP

c) 5 moles of $O_2$

1 mole of $O_2$ contains $6.023\times 10^{23}$ molecules at STP

5 moles of  $O_2$ contains= $\frac{6.023\times 10^{23}}{1}\times 5=30.12\times 10^{23}$ molecules at STP

d) $5\times 10^{23}$ molecules of $CO_2(g$