Question

10 g of NaOH was added to 200cc of N/2(f=1.5) H2SO4 the volume was diluted to two liters calculate resulting molarity of dilute solution ?

Answer 1

Given weight of NaOH= 10g

Volume of H2SO4=200cc

Normality of H2SO4= (1/5)*1.5=0.75

Number of gram equivalence of NaOH= (Weight/ Equivalent weight)= (10/40)=0.25

No. of gram equivalence of H2SO4=

(Normality* volume in Lt) = (0.75*0.2) =0.15

Hence it is basic.

Now, remaining no of gram equivalence of NaOH after it neutralises H2SO4= 0.25-0.15= 0.1

Normality of diluted remaining base=(no. of gram equivalence of remaining base)/ total diluted volume in litre

= (0.1/2)

= 0.05

Finally,

Molarity= normality/ acidity

= 0.05/1

= 0.05 M

Hence the molarity of the dilute solution is 0.05 M

I hope you got it!

Answer 2

Answer:

.05

Explanation:

no of gram equivalent of naoh = 10/40

= .25

no of gram equivalent wt. of H2SO4= N* volume in L = (1/2*1.5)*(200/1000) [normality = normality* normality factor]

= 0.15

so, the solution is basic

number of unreacted gram of NAOH = 0.25-0.15

= 0.1

normality of soln. is = 0.1/2 = 0.05N

molarity= normality* basicity= 0.05*1 = 0.05M