Question

$1.51\times {10}^{23}{Ca}^{2+}$ ions =______________g

Answer 1

As we know that,

1 mole of ${Ca}^{+2}$ ions $=6.023\times {10}^{23}{Ca}^{+2}\text{ions}$

$\therefore$ No. of moles in $1.51\times {10}^{23}{Ca}^{+2}\text{ions}=\frac{1}{6.023\times {10}^{23}}\times 1.51\times {10}^{23}=0.25\text{mole}$

Mass of $1$ mole ${Ca}^{+2}$ ions $=40g$

Mass of $0.25$ mole of ${Ca}^{+2}$ ions $=0.25\times 40=10g$

Hence, $1.51\times {10}^{23}{Ca}^{+2}$ ions $=10g$