Question

4.4 g of $C{O}_2$ and 2.24 litre of $H_2$ at STP are mixed in a container. The total number of molecules present in the container will be:

• A. $6.022\times {10}^{23}$
• B. $1.2044\times {10}^{23}$
• C. $6.023\times {10}^{26}$
• D. $6.023\times {10}^{24}$

Answer 1

Answer: (B)

$1.2044\times {10}^{23}$

No.of moles of $C{O}_2$ in 4.4 g $C{O}_2=4.4/44=0.1$ mole.

No.of moles of $H_2$ gas in 2.24 liters at STP=2.24/22.4= 0.1 mole.

So total moles of $H_2$ and $C{O}_2$ gas are =(0.1+0.1)=0.2 moles.

Total number of Molecules in the container$=0.2\times 6.022\times {10}^{23}=1.2044\times {10}^{23}$