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

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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.