How many L of a 4 M KCL solution are necessary to obtain 7.46 g of KCl

-How many L of a 4 M KCL solution are necessary to obtain 7.46 g of KCl?

-How many mL of original solution are needed to prep 20 mL of a .25 M KNO3 from a 6 M KNO3 solution?

-I have 100 mL of a .35 M KBr solution. I need a .400 M solution. Assuming no change in volume. Ho many moles of KBr need to be ADDED to the solution to increase the concentration to .400 M?

1) 7.46 g KCl / (74.5515 g/mol) = .100 mol KCl
since M = mol/L, you can set up an equation:
(4 mol KCl/L) * (x L) = .100 mol KCl
x = .0250 L solution

2) 20 mL = .020 L
(.25 mol KNO3/L) * (.020 L) = .0050 mol KNO3 needed
(6 mol KNO3/ L) * (x L) = .0050 mol KNO3
x = 8.3 * 10^-4 L
=.83 mL

3) 100 mL = .100 L
(.35 mol KBr/L) * (.100 L) = .035 mol KBr
(.400 mol KBr/L) * (.100 L) = .040 mol KBr
.040 mol KBr-.035 mol KBr = .005 mol KBr (that you need to add)

How many L of a 4 M KCL solution are necessary to obtain 7.46 g of KCl?

Can conver this to grams/liter to make it easy. 4 mole/liter x 74.55 grams/mole = 298.2 grams/liter
So: 7.46 / X = 298.2 grams/liter X = 0.025 liters

How many mL of original solution are needed to prep 20 mL of a .25 M KNO3 from a 6 M KNO3 solution?

Calculate the number of moles in 20 mL of a .25 M KNO3: 0.25 x 20/1000 = 0.005 moles
So, we need 0.005 moles in X mL of 6 M KNO3 so in this case divide: 0.005 / 6 x 1000 = 0.83 mL

I have 100 mL of a .35 M KBr solution. I need a .400 M solution. Assuming no change in volume. Ho many moles of KBr need to be ADDED to the solution to increase the concentration to .400 M?

Calculate the number of moles in the first solution: 0.35 x 100/1000 = 0.035 mles
Calculate the number of moles in the desired solution: 0.4 x 100/1000 = 0.04 mles
The difference is 0.04 mles - 0.035 mles = 0.005 mles
Multiply by the molecular weight to get grams: 0.005 mles x 119 = 0.60 grams.
So add 0.60 grams.