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 ?
Answers
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:
.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