34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfuric acid solution?

0.1045M is the concentration of the original sulfuric acid solution

Explanation:

Titration is done to know the volume or concentration of unknown electrolyte.

Data given:

volume of acid = 50 ml

molarity of acid =?

volume of base NaOH = 34.62 ml

molarity of the base = 0.1510

For titration the formula used is,

Macid x Vacid = Mbase  x V base

Putting the values in above equation:

M acid x 50 = 34.62 x 0.1510

Macid =

           = 0.1045 M is the molarity of sulphuric acid solution used in neutralization of 0.1510 M base solution.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o