How many molecules of water of hydration are present in 252mg of oxalic acid $(H_{2} C_{2} O_{4} . 2 H_{2} O)$ ?

2.68 \times 10^{18}

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2.52 \times 102^{21}

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1.83 \times 10^{24}

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2.4 \times 10^{21}

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Solution

The correct option is D $2.4 \times 10^{21}$
Given, mass of oxalic acid = 252 mg or 0.252g

molar mass of hydrated oxalic acid = 126

so, number of moles of hydrated oxalic acid = $\frac{0.252}{126} = 0.002$ moles

1 mole of hydrated oxalic acid contains 2 moles of water molecules

so, 0.002 moles of hydrated oxalic acid contains 0.004 moles of water

1 mole of water have number of molecules = $6.023 \times 10^{23}$

so, molecules of water present in 0.004 moles = $6.023 \times 10^{23} \times 0.004 = 2.4 \times 10^{21}$

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