The molarity of a 0.2 N `Na_(2)CO_(3)` solution will be :

Options

A. `0.1M`
B. `0M`
C. `0.4M`
D. `0.2`

Answer

A

Explanation

`"Molarity=normality"xx("equivalent weight")/("molecular weight")`
`"Given,normality of "Na_(2)CO_(3)" solution"=0.2N`
Equivalent weight =M
Moleular weight = 2M ` " "(.:'Na_(2)CO_(3)" is depositive")`
`:."Molarity"=0.2xx(M)/(2M)=0.1M`