Question

How many molecules of water of hydration are present in 630mg of oxalic acid (H2C2O4.2H2O)

Answer 1

Number of molecules available in oxalic acid is equal to 630 divided by 126 which come to 5.

 Number of molecules of water present is equal to 5 x 6.02 x 10 ^ 23

Therefore, number of molecules of water of hydration present in 630 mg oxalic acid is 30.1 x 10^ 23.

Answer 2

Explanation:

molar mass of H2C2O4.2H2O=126 g

no. of molecules of water in one molecule of oxalic acid = 2 * 6.022 * 10^23

no. of water molecules in 630*10^-3 g of oxalic acid=

2 * 6.022 *10^23 * 630 * 10^-3

/ 126 = 6.022 * 10^21