1.

The mole fraction of oxalic acid (Molar mass 63) required to prepare 0.10 m solution in water is

Answer»

1
0.0018
6.3
0.0992

Solution :`X_(2)= (mM_(1))/(1000+mM_(1))`
`=(0.1 XX 18)/(1000 + 0.1 xx 18)`
=0.0018


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