10 millimoles of a di acidic base exactly neutralises 100ml of an acid. Then the Normality of that acid is

Given,
Number of moles of base is equal to 10 millimoles =$$10\times {10}^{-3}$$mol

The acidity of the base = 2.

The number of equivalence of base will be = number of moles of base X acidity of base

                                                                           =$10\times {10}^{-3}\times 2=2\times {10}^{-2}$

Given the diacidic base has completely neutralized the base. 

Therefore, the number of equivalents of base =the number of equivalences of acid.

The volume of acid used for neutralization is100mL. The volume of the acid in liters will be

$\mathrm{V}=\frac{100}{1000}=0.1\mathrm{L}$
Hence the normality of the acid = $\frac{\mathrm{equivalence} \mathrm{of} \mathrm{acid}}{\mathrm{volume} \mathrm{of} \mathrm{acid}}$

$\mathrm{N}=\frac{2\times {10}^{-2}}{0.1}=0.2\mathrm{N}$