Question

50 cc of decinormal NaOH solution will be completely
neutralised by 'x' ml of decimolar H, SO, solution. The
value of 'x' is
1) 1 2) 10
3) 25 4) 100​

Answer 1

Explanation:

molecular mass of NaOH = 40 g

molarity of stock solution = 40×5002×1000 = 0.1 M

no. of moles of NaOH in 40.50 ml  of solution = 0.1×100040.5 = 0.004504 moles

  

for neutralisation, we have: no. of moles of HCl = no. of moles of NaOH

1 decimolar = 0.1 M

 so, number of moles of HCl in 10 ml decimolar solution = 100010×0.1 = 0.001

 number of moles of HCl in 20 ml decimolar solution = 100020×0.1 = 0.002

 number of moles of HCl in 50 ml decimolar solution = 100050×0.1 = 0.005