in 252 mg of oxalic18) How many molecules of water of hydration are pr

1.	in 252 mg of oxalic18) How many molecules of water of hydration are presentacid?
Answer» Number of molecules available in oxalic acid is equal to 252 divided by 126 which come to 2. Number of molecules of water present is equal to 2 x 6.02 x 10 ^ 23 Therefore, number of molecules of water of hydration present in 252mg oxalic acid is 12.04 x 10^ 23.

in 252 mg of oxalic18) How many molecules of water of hydration are presentacid?

Answer»

Number of molecules available in oxalic acid is equal to 252 divided by 126 which come to 2.

Number of molecules of water present is equal to 2 x 6.02 x 10 ^ 23

Therefore, number of molecules of water of hydration present in 252mg oxalic acid is 12.04 x 10^ 23.