The mass of $1 \times 10^{22}$ molecule of blue vitriol $({C u S O}_{4} . x H_{2} O)$ is 4.144 g, the value of $x$ will be:

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Solution

Mass of $1 \times 10^{22} C u S O_{4} . x H_{2} O = 4.144 g$
$∴$ Mass of $6.022 \times 10^{23} C u S O_{4} . x H_{2} O = \frac{4.144 \times 6.022 \times 10^{23}}{1 \times 10^{22}} = 249.55 g$
$∴ 1$ mole of $C u S O_{4} . x H_{2} O = 249.55 g$
$⟶ C u S O_{4} = 159.6 g / m o l$
$∴ x H_{2} O = 249.55 - 159.6 = 89.95 g ≃ 90 g$
$1$ mole of $H_{2} O = 18 g$
$∴ x = 90 / 18 = 5$

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