Answer to Question #79084 in Inorganic Chemistry for Krishna

Question

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

Answer 1

NA=6.02*10^23
m=630 mg= 0.63 g.
Mr=(2*1)+(12*2)+(16*4)+(2*18)=126 g/mol.
N molecules of water=?
Formula:
n=m/Mr n= 0.63/126=0.005
N molecules of water=n* NA
N molecules of water= 0.005*6.02*10^23=3.01*10^21
Answer: 3.01*10^21