Question

25 mL 0.1 N H2S04 neutralized with 20 mL xN Na2CO3. What will be the g/liter of Na2CO3? (M.W Na2CO3 = 106 g/mol)
1) 8.48 g
4) 13.25 g
2)4.24g
3) 6.625 g​

Answer 1

Answer:

6.36g

Explanation:

N1V1=N2V2

25*0.1=20*x

x=012N

now,0.12=mass of Na2CO3/eq. mass*volume

      0.12=mass/53*V

mass /litre=0.12*5

                 =6.36g