The molar mass of \({}^{7}_{3}\mathrm{Li}\) is 7 g. Taking the Avogadro number as \(6\times10^{23}\ \mathrm{mol^{-1}}\), the number of neutrons in 7 g of \({}^{7}_{3}\mathrm{Li}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2.4\times10^{24}\)
A nucleus \( {}^{A}_{Z}\mathrm{X} \) has \( Z \) protons and \( A-Z \) neutrons. Here 7 g of lithium-7 is one mole, so it contains \( 6\times10^{23} \) nuclei.
\( \begin{aligned} N &= A-Z \\ &= 7-3 \\ &= 4 \end{aligned} \)
Each nucleus has 4 neutrons. For one mole of nuclei,
\( \begin{aligned} \text{Number of neutrons} &= 4\times N_A \\ &= 4\times6\times10^{23} \\ &= 2.4\times10^{24} \end{aligned} \)
NCERT (2026-27):
Atomic Masses and Composition of Nucleus – “The composition of a nucleus can now be described using the following terms and symbols:”
Z - atomic number = number of protons\( \qquad [13.4(a)] \)
N - neutron number = number of neutrons\( \qquad [13.4(b)] \)
A - mass number = Z + N = total number of protons and neutrons\( \qquad [13.4(c)] \)