Question

Determine the molarity of a solution formed by dissolving 468 mg of mgi2 in enough water to yield 50.0 ml of solution.

Answer 1

Answer : The molarity of a solution is, 0.0337 M

Explanation : Given,

Mass of $MgI_2$ = 468 mg  = 0.468 g   (1 mg = 0.001 g)

Volume of solution = 50.0 mL

Molar mass of $MgI_2$ = 278 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }MgI_2\times 1000}{\text{Molar mass of }MgI_2\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.468g\times 1000}{278g/mole\times 50.0mL}=0.0337mole/L=0.0337M$

Therefore, the molarity of a solution is, 0.0337 M

Answer 2

The molarity of the solution formed is 0.0337 M

From the question,

We are to determine the molarity of the solution formed.

First, we will determine the number of moles MgI₂ dissolved

Mass of MgI₂ dissolved = 468 mg = 0.468 g

From the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of MgI₂ = 278.1139 g/mol

∴ Number of moles of MgI₂ present = $\frac{0.468}{278.1139}$

Number of moles of MgI₂ present = 0.001682764 mole

Now,

For the molarity of the solution formed,

Using the formula

$Molarity = \frac{Number\ of\ moles}{Volume}$

Volume of the solution = 50.0 mL = 0.050 L

∴ Molarity of the solution = $\frac{0.001682764}{0.050}$

Molarity of the solution = 0.033655

Molarity of the solution ≅ 0.0337 M

Hence, the molarity of the solution formed is 0.0337 M

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