How many grams of KCl are necessary to prepare 1.5 liters of a 0.500-M solution of KCl?

Answer 1

mols\L=M

x\2.00=0.500

x=.500(2.00)

x=1.00 mol

K=39.10 g/mol

N=14.01 g/mol

O=16.00 g/mol

39.10+14.01+3(16.00)=101.11g/mol

1.00mol(101.11g/mol) = 101.11g of KNO3

Answer 2

Consider the formula for molarity. M = n solute / V solution L Isolate the mass in the formula. M = (mass / molar mass) / V solution L MV = mass / molar mass MV(molar) = mass Solve for the mass. (0.500 M)(1.50 L)(74.55 g/mol) = mass 55.9 g = mass (Answer)

Answer 3

.500L x 1.5mol/L x 101.41g/mol= 76g potassium nitrate.

Answer 4

The answer is 4,9 moles; but this solubility is possible only above 25 0C.

Answer 5

101 g / mole x ( 2.0 mole / liter) x (.5 L) = 101 grams

Answer 6

Any amount - if you have a sufficient amount of solvent.

Answer 7

Assuming you mean 2 M solution and not 2 m (molal), you would do it as follows:
2.0 moles/L x 1.5 liters = 3.0 moles KNO3 must be dissolved in 1.5 liters.

Answer 8

1.5 x 2 = 3 moles total

Answer 9

We need 10,11 g potassium nitrate.

Answer 10

The answer is 3 moles HNO3.