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

(d) 5^1023 molecules of co2g

Answer 2

Answer : The correct option is, (B) 5 liter of $CH_4(g)$

Solution :

First we have to calculate the number of molecules in 5 liter of $NO_2(g)$

As, 22.4 L contains $6.022\times 10^{23}$ number of molecules

So, 5 L contains $\frac{5L}{22.4L}\times 6.022\times 10^{23}=1.344\times 10^{23}$ number of molecules

(a) Now we have to calculate the number of molecules in 5 gram of $H_2(g)$

As, 2 g contains $6.022\times 10^{23}$ number of molecules

So, 5 g contains $\frac{5g}{2g}\times 6.022\times 10^{23}=15.055\times 10^{23}$ number of molecules

(b) Now we have to calculate the number of molecules in 5 liter of $CH_4(g)$

As, 22.4 L contains $6.022\times 10^{23}$ number of molecules

So, 5 L contains $\frac{5L}{22.4L}\times 6.022\times 10^{23}=1.344\times 10^{23}$ number of molecules

(c) Now we have to calculate the number of molecules in 5 moles of $O_2(g)$

As, 1 mole contains $6.022\times 10^{23}$ number of molecules

So, 5 moles contains $5moles\times 6.022\times 10^{23}=30.11\times 10^{23}$ number of molecules

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

Hence, the correct option is, (B) 5 liter of $CH_4(g)$