How many grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water?

a. 83.9

b. 0.500

c. 1.13

d. 2.00

Answer: Option A) 83.9g

Explanation:

KCl is the chemical formula of potassium chloride.

Given that,

Amount of moles of KCl (n) = ?

Volume of KCl solution (v) = 0.75L

Concentration of KCl solution (c) = 1.5M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 1.5M x 0.75L

n = 1.125 mole

Now given that,

Amount of moles of KCl (n) = 1.125

Mass of KCl in grams = ?

For molar mass of KCl, use the molar masses of:

Potassium, K = 39g;

Chlorine, Cl = 35.5g

KCl = (39g + 35.5g)

= 74.5g/mol

Since, amount of moles = mass in grams / molar mass

1.125 mole = m / 74.5g/mol

m = 1.125 mole x 74.5g/mol

m = 83.81g

Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water