Question

the no.of molecule at ntp in 1ml of an ideal gas will be

Answer 1

22400 ml of ideal gas= 6.022× 10²³= Na particles
So 1 ml of gas = 6.022×10²³÷22400
=2.6889×10(power 19) molecules

Answer 2

By ideal gas law, 22.4 litres of a gas or 22400 ml of a gas contains Avogadro’s number of molecules $=6.022 \times 10^{23}$ molecules.

Therefore, at normal temperature and pressure (NTP), 1 ml of the gas will contain  

$=\frac{\left(6.022 \times 10^{23}\right)}{22400} \text { molecules }=2.688 \times 10^{19}\ molecules$.

Therefore, 1 ml of an ideal gas consists of $\bold{2.688 \times 10^{19}}$ molecules of the gas.

Normal Temperature and Pressure (NTP) is considered to be a gas at 20 degree Celsius and 1 atmospheric pressure.