Question

How many molecules of water of hydration are present in 252mg of oxalic acid $(H_2{C}_2{O}_4.2H_{2}O)$?

• A. $2.68\times {10}^{18}$
• B. $2.52\times {102}^{21}$
• C. $1.83\times {10}^{24}$
• D. $2.4\times {10}^{21}$

Answer 1

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 = $6.023\times {10}^{23}\times 0.004=2.4\times {10}^{21}$