Question

What is the molar concentration (in moles/L) of a solution made with 3.744 g of Mg(NO3)2 dissolved in enough water to make 50.0 mL of solution?

Answer 1

Answer:

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$

Explanation:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

In this case:

• Mg: 24.3 g/mole
• N: 14 g/mole
• O: 16 g/mole

So, the molar mass of Mg(NO₃)₂ is:

Mg(NO₃)₂= 24.3 g/mole + 2*(14 g/mole + 3*16 g/mole)= 148.3 g/mole

So, if you have 3.744 g of Mg(NO₃)₂, you can apply the following rule of three: if 148.3 grams of Mg(NO₃)₂ are present in 1 mole, 3.744 grams in how many moles are present?

$moles=\frac{3.744 grams*1mole}{148.3 grams}$

moles= 0.025

Then you have:

• number of moles=0.025
• volume= 50 mL= 0.05 L (being 1,000 mL= 1 L)

Replacing in the definition of molarity:

$molarity=\frac{0.025 moles}{0.05 L}$

you get:

$molarity=0.5 \frac{moles}{L}$

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is $0.5 \frac{moles}{L}$