Question

Consider the following nuclear reactions involving $X$ and $Y$.
$X\to Y{+}_2^{4}He$
$Y{\to }_8^{18}O{+}_1^{1}H$
If both neutrons as well as protons in both the sides are conserved in nuclear reaction then moles of neutrons in 4.6 gm of $X$ is:

• A. $2.4N_A$
• B. $2.4$
• C. $4.6$
• D. $0.2N_A$

Answer 1

The correct option is B $2.4$

If both neutrons, as well as protons in both the sides, are conserved in nuclear reaction then the reactions will be written as:

${}_{11}^{23}Na{\to }_9^{19}F{+}_2^{4}He$

${}_9^{19}F{\to }_8^{18}O{+}_1^{1}H$

$X$ is $Na$ which has mass number $A=23$.

Number of protons + Number of neutrons = Mass number A = 23

Number of neutrons = 23 - number of protons $=23-11=12$

1 mole $Na=23$ gm $Na=12$ moles of neutrons

4.6 gm $Na=\frac{(4.6\times 12)}{23}=2.4$ moles of neutrons