Question

If we have a pure $N{a}_{2}S{O}_4.5H_{2}O$ sample and it contains $2.2\times {10}^{10}$ electrons, the mass of the sample is $x\times {10}^{-14}g$. x is

Answer 1

Number of electrons in 1 molecule of $N{a}_{2}S{O}_4.5H_{2}O=120$
$120\times 6.022\times {10}^{23}$ electrons are present in 1 mole of $N{a}_{2}S{O}_4.5H_{2}O$
$2.2\times {10}^{10}$ electrons will be present in $\frac{2.2\times {10}^{10}}{120\times 6.022\times {10}^{23}}$ moles of $N{a}_{2}S{O}_4.5H_{2}O$
$=3.04\times {10}^{-16}$ moles of $N{a}_{2}S{O}_4.5H_{2}O$
Mass of $N{a}_{2}S{O}_4.5H_{2}O=\text{no.of moles}\times \text{molar mass}$
$=3.04\times {10}^{-16}\times 232=7.05\times {10}^{-14}\text{g}$