Question

What volume of water is used to prepare a 3.5 M solution if 45 grams of Na2CO3 is used?

Answer 1

121.3 ml

Step-by-step explanation:

Mole wts

Na 22.989 x 2

C 12.011

O 15.999 x 3 total = 105.986

45 grams is 45/105.986 of a mole

45/105.986 in one liter would be .424584 M

.424585 mole / x liter = 3.5

x = .1213 liter = 121.3 ml