An aqueous solution of 6.3 g of oxalic acid dehydrate is made up to 250 ml. the volume of 0.1 m naoh required to completely neutralise 10 ml of solution is 20 ml 4ml 10 ml 40 ml

A) 50 g of iron b) 5 moles of N2 c)0.1 g atom of Ag d)10^23 atoms of carbon

Dear student,
moles of oxalic acid = 6.3 / 126= 0.05
molarity of oxalic acid solution = 0.05  * 1000 / 250 = 0.20 m
now, molarity of naoh solution = 0.1 m (same as normality )
    since  m1v1  = m2v2
      0.20* 10 = 0.1 * v2 
v2  = 20  ml
so, 20 ml of 0.1n  naoh  is needed 

HEY FOLK...

Molecular formula of Oxalic acid dehydrate = H2C2O4 . 2H2O

Molecular mass of Oxalic acid dehydrate = 126

Since it is dibasic acid so,

equivalent weight = moleculas mass / 2

                         = 126/2

                         = 63

Normality = given mass / eq. wt. × volume ( litre )

               = 6.3/63 × 1000/250

               = 0.4 N

we know that,

 N1V1 = N2V2

 0.1 ×V1 = 0.4 × 10

Therefore, V1 = 40 mL.

HOPE IT HELPS...

#Mme