Asample of 0.8214 g of kmno4
was dissolved in water and made up to the volume in a 500 ml volumetric flask. a 3.000 ml sample of this
solutions was transferred to a 100 ml volumetric flask and diluted to the mark with water. next, 10.00 ml of the
diluted solution were transferred to a 50 ml volumetric flask and diluted to the mark with water. calculate the
concentration (in molarity) of the final solution.

i think a, hope it .

explanation:

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