Question

What is the molarity of a solution containing 78.6 g of mgcl2 dissolved in 1.00 l of solution?

Answer 1

Molar mass MgCl2 = 95.211 g/mol

number of moles:

1 mole ---------- 95.211 g
? moles --------- 78.6 g

78.6 x 1 / 95.211 => 0.8255 moles

Therefore:

M = n / V

M = 0.8255 / 1.00

M = 0.8255 mol/L

Answer 2

Answer: Molarity of the solution is 0.83 M

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`Molarity=(n)/(V_s)`

where,

n= moles of solute

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}=(78.6g)/(95g/mol)=0.83mole`

`V_s` = volume of solution in L

`Molarity=(0.83moles)/(1L)=0.83M`

Thus the molarity of the solution is 0.83 M