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Question: At STP, volume of a gas in a discharge tube is \[1.12 \times {10^{ - 4}}ml\] . Number of molecules p...

At STP, volume of a gas in a discharge tube is 1.12×10−4ml1.12 \times {10^{ - 4}}ml . Number of molecules present in the gas are:
A. 3×10233 \times {10^{23}}
B. 3×10203 \times {10^{20}}
C. 3×10153 \times {10^{15}}
D. 3×10193 \times {10^{19}}

Explanation

Solution

Volume of gas is given at STP i.e. 22400ml, the number of molecules it has is equal to the Avogadro’s number. So, using a unitary method, we can find out the number of molecules present in a gas at any volume of a gas at standard temperature and pressure.

Complete step by step answer:
Gram molecular volume is defined as the volume bound up by a mole of each gas under standard conditions such as 273K temperature and one bar pressure. Its value was found to be the same for all gases i.e. 22.4 litres and it is also known as molar volume and at standard temperature and pressure (STP).
We are provided that at STP, the volume of gas in a discharge tube is 1.12×10−4ml1.12 \times {10^{ - 4}}ml
And we know that one mole of an ideal gas occupies 22400ml or 22.4 litres of volume at STP. So, we can say that at STP, 22.4 litres of volume contain 6.023×10236.023 \times {10^{23}} number of molecules.
Therefore, number of molecules present in 1.12×10−4ml1.12 \times {10^{ - 4}}ml will be
=1.12×10−4ml×6.023×102322400ml=3×1015= \dfrac{{1.12 \times {{10}^{ - 4}}ml \times 6.023 \times {{10}^{23}}}}{{22400ml}} = 3 \times {10^{15}}

Hence, the correct option is (C).

Additional information: As per Avogadro’s law, the ratio of volume and amount of gaseous substance is constant at constant pressure and temperature. The value of this constant (k) can be determined using the following equation:
k=(RT)/Pk = \left( {RT} \right)/P
Under standard conditions for temperature and pressure, the value of T corresponds to 273.15 Kelvin and the value of P corresponds to 101.325 kiloPascals. Therefore, the volume occupied by one mole of a gas at STP is:   (8.314J.mol−1.K−1)×(273.15K)/(101.325kPa)=22.4litres\;\left( {8.314J.mo{l^{ - 1}}.{K^{ - 1}}} \right) \times \left( {273.15K} \right)/\left( {101.325kPa} \right) = 22.4litres
That is why, we say that one mole of any gaseous substance occupies 22.4 litres of volume at STP.

Note:
For solving such questions, make sure that the volume is in litres or millilitres because the volume of one mole of a gas at STP and NTP is in litres i.e., 22.4 litres. At standard temperature and pressure (STP), the pressure is taken as 1 bar and temperature as 273K whereas at normal temperature and pressure (NTP), the pressure is 1 atm and temperature is 273K.