Question

Volume of a gas at NTP is $ 1.12\times {{10}^{-7}}c{{m}^{3}} $ . The number of molecules in it is

• A. $ 3.01\times {{10}^{12}} $
• B. $ 3.01\times {{10}^{24}} $
• C. $ 3.01\times {{10}^{23}} $
• D. $ 3.01\times {{10}^{20}} $

Answer 1

Answer: (A)

Given, $V=1.12 \times 10^{-7} cm ^{3}$
$\because 22400\, cm ^{3}$ of the gas at STP has molecules
$=6.02 \times 10^{23}$
$\therefore 1.12 \times 10^{-7} cm ^{3}$ of the gas at STP will have molecules
$=\frac{6.02 \times 10^{23}}{22400} \times 1.12 \times 10^{-7} $
$=3.01 \times 10^{12} $ molecules