How many grams of a dibasic acid (mol. Mass `200`) should be present in `100 mL` of the aqueous solution to give `0.1 N` solution.

Options

A. `1 g`
B. `20 g`
C. `10 g`
D. `2 g`

Answer

1

Explanation

Normality (N) `=(n_(eqs)(solute))/(V_(mL)(solute))xx(1000 mL)/L …(1)`
`n_(eqs)(solute)=(mass (solute))/("gram eq mass "(solute))`
Equivalent mass (acid)
`=("Formula mass acid")/("Basicity of acid")`
Thus, gram eq. mass (acd) `=100" g "eq^(-1)`
Using these resulys in eq (1), we have
`0.1" eq " L^(-1)=(mass (solute)//100 g eq^(1))/(100 mL)xx(1000 mL)/L`
or mass `(solute)=1 g`