Question

What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?

Answer 1

0.587 M is the molarity of a solution.

Step-by-step explanation:

`Molarity=\frac{\text{mass of solute}}{\text{Molar mass of solute}* \text{Volume of solution (L)}}`

Mass of magnesium chloride = 55.8 g

Molar mass of magnesium chloride = 95 g/mol

Volume of the solution = 1.00 L

Molarity of the solution :

`=(55.8 g)/(95 g/mol* 1.00 L)=0.587 M`

0.587 M is the molarity of a solution.

Answer 2

The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M