What is the molarity of a solution made by dissolving 58.8 of NaOH in 3.1 L of
solution?

The molarity of the solution is 0,47 M

Explanation:

Molarity is a concentration measurement that expresses the moles of solute (in this case NaOH) in 1 liter of solution (1000ml). First we calculate the mass of 1 mol of NaOH, to calculate the moles in 58.8 g of said compound:

Weight  1 mol NaOH= Weight Na + Weight O + Weight H=23 g + 16 g +1 g

Weight 1 mol NaOH= 40 g/mol

40 g ---1 mol NaOH

58,8g---x= (58,8g x 1 mol NaOH)/40g =1,47 mol NaOH

3,1 L solution 1,47 mol NaOH

1 L  solution x= (1 L  solution x 1,47 mol NaOH)/3,1 L solution

x= 0,47 mol NaOH ---> The molarity of the solution is 0,47 M

wher is the rest of 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