Calculate the morality of each of the following solutions: a. 15.4 g KCl in 289.2 mL solution b. 14.4 g of CaCl2 in 0.614 L solution c. 28.0 mL of 3.00 M H2SO4 diluted to 0.250 L

Answers

Answer 1

Answer: The answer is:
a. 0.712 M
b. 0.210 M
c. 0.336 M

Molarity is a measure of the concentration of solute in a solution.
It can be expressed as moles of solute ÷ volume of solution:
c = n ÷V
where:
c - concentration of solute,
n - moles of solute
V - volume of solution

n can be expressed as:
n = m ÷ Mr
where:
n - moles of solute
m - mass of solute
Mr - relative molecular mass

a. We know volume:
V = 289.2 mL = 0.2892 L
We need n and c.

n = m ÷ Mr
m = 15.4 g
Mr (KCl) = 74.55 g/mol
n = 15.4 g ÷ 74.55 g/mol
n = 0.206 mol

Thus,
c = 0.206 mol ÷ 0.2892 L
c = 0.712 mol/L = 0.712 M

b. We know volume:
V = 0.614 L
We need n and c.

n = m ÷ Mr
m = 14.4 g
Mr (CaCl₂) = 110.98 g/mol
n = 14.4 g ÷ 110.98 g/mol
n = 0.129 mol

Thus,
c = 0.129 mol ÷ 0.614 L
c = 0.210 mol/L = 0.210 M

c. We can use formula:
m₁V₁ = m₂V₂
m₁ = 3 M
V₁ = 28 mL= 0.028 L
m₂ = ?
V₂ = 0.250 L
Thus:
3 M × 0.028 L = m₂× 0.250 L
m₂ = 0.336 M

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