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