How many moles of NaCl are present in a solution with a molarity of 8.59M 1 point

and 125 mL of solution? 1L = 1000 mL

Answer : The number of moles of NaCl present in a solution are, 1.07 moles.

Explanation : Given,

Molarity of = 8.59 M

Volume of solution = 125 mL  = 0.125 L      (1L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

Now put all the given values in this formula, we get:

Therefore, the number of moles of NaCl present in a solution are, 1.07 moles.

law of conservation of matter

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