How many molecules of water of hydration are present in 630mg of oxalic acid (H2C2O4.2H2O)
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Answered by
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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.
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.
Answered by
3
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
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