Calculate the molecular weight of 0.2g H3X solution if 50ml of this acid neutralized with 40ml of 0.03N Ca(oH)2 then calculate strength of this solution in ppm

1. Molecular weight = 333.3 g/mol

2. Strenght = 2222.2 mg/L

Explanation:

First, we need to find molarity (M) from normality (N) as follows:

Where n is the number of OH⁻ = 2

Now, when the H₃X solution is neutralized with Ca(OH)₂ we have:

Hence, the molecular weight (MW) of H₃X is:    

Finally, we can calculate the strength (S) of this solution in ppm as follows:

I hope it helps you!

the right answer is c i just did the question.

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