Question

The number of nitrogen molecules present in 1c.c of gas at NTP is:
(a)- $2.67\text{ x 1}{{\text{0}}^{22}}$
(b)- $2.67\text{ x 1}{{\text{0}}^{21}}$
(c)- $2.67\text{ x 1}{{\text{0}}^{20}}$
(d)- $2.67\text{ x 1}{{\text{0}}^{19}}$

Answer 1

The number of molecules of nitrogen can be calculated by multiplying the number of moles of the nitrogen to the Avogadro’s number. The number of moles of the gas is calculated by dividing the given volume to the volume of one mole of nitrogen gas at NTP.

Complete answer:
First, we have to calculate the number of moles of nitrogen gas.
The number of moles of the gas is calculated by dividing the given volume to the volume of one mole of nitrogen gas at NTP.
$\text{Number of moles = }\dfrac{\text{Given volume }}{\text{volume of the gas at NTP}}$
The given volume of the gas is 1c.c (cubic centimeter). So, we have to convert it into liters. For converting it into liters divide the given value with 1000. We get,
$1\text{ c}\text{.c = 1}{{\text{0}}^{-3}}\text{ L}$
According to the NTP (normal temperature and pressure), the volume of one mole of gas is 22.4L.
The volume of the nitrogen at NTP (normal temperature and pressure) is 22.4 L.
So, the number of moles will be,
$\text{Number of moles = }\dfrac{\text{Given volume }}{\text{volume of the gas at NTP}}=\dfrac{\text{1}{{\text{0}}^{-3}}}{22.4}$
Now, for calculating the number of molecules of nitrogen can be calculated by multiplying the number of moles of the nitrogen to the Avogadro’s number.
Avogadro’s number may be defined as the number of atoms present in one gram atom or one mole of the element, or the number of molecules present in one gram molecule or one mole molecule of the substance. The value of Avogadro’s number is $6.022\text{ x 1}{{\text{0}}^{23}}$
$Avogadro's\ number\text{ (}{{\text{N}}_{A}}\text{) = 6}\text{.022 x 1}{{\text{0}}^{23}}$
So, $number\text{ }of\text{ }nitrogen\text{ }molecules\text{ }=\text{ n x 6}\text{.023 x 1}{{\text{0}}^{23}}$
$=\text{ }\dfrac{{{10}^{-3}}}{22.4}\text{ x 6}\text{.023 x 1}{{\text{0}}^{23}}$
$=\ \text{2}\text{.67 x 1}{{\text{0}}^{19}}$

Hence, the correct answer is an option (d)- $\ \text{2}\text{.67 x 1}{{\text{0}}^{19}}$

Note:
The calculation must be done correctly when options are given very similarly. The volume should always be converted into liters because the volume of one mole of a gas at STP or NTP is in liters. The volume STP and NTP is 22.4 L. At STP (standard temperature and pressure), the pressure is 1 bar and temperature is 273 K. At NTP (normal temperature and pressure), the pressure is 1 atmosphere and temperature is 237 K.