In the laboratory you are asked to make a 0.112 m magnesium nitrate solution using 355 grams of water. How many grams of magnesium nitrate should you add?

Question

In the laboratory you are asked to make a 0.112 m magnesium nitrate solution using 355 grams of water.

How many grams of magnesium nitrate should you add?

Answer 1

0.112 m means 0.112 molal which means 0.112 moles of Mg(NO₃)₂ per kg of solvent.

Since we have 355 g of solvent, this is equal to 0.355 kg.

The amount of Mg(NO₃)₂ needed is thus calculated as 0.112 mols/kg x 0.355 kg = 0.03976 mols

molar mass Mg(NO₃)₂ = 148.3 g/mol

mass Mg(NO₃₎₂ needed = 0.03976 mols x 148.3 g/mol = 5.90 g (3 sig. figs.)

Answer 2

Hello, Tee,

The definition of M [capital M is an abbreviation for molar, m could be confused with meters] is moles per liter. We want a 0.112M solution, which means 0.112 moles of magnesium nitrate per 1 liter of solvent (water). We are only given 255 grams of water, so we'll need to scale down the moles of Mg(NO₃)₂.

We need to do a couple of steps to get an answer. First, we need the molar mass of Mg(NO₃)₂. I find 148 grams/mole. If we were making 1 liter of a 1 M Mg(NO₃)₂ solution, we'd add 148 grams of Mg(NO₃)₂ and add water to make 1 liter.

Next, we need to scale down the mass we will use due to both the lower concentration and due to the smaller volume. We'll assume water has a density of 1g/ml, so the final solution volume will be 0.255 liters.

We want a 0.112M solution. Find the moles of Mg(NO₃)₂ needed for that by multiplying the molarity times the target volume, 0.225 liters:

(0.12 moles/liter)*(0.225 liter) = 0.02856 moles Mg(NO₃)₂ . Convert this value of moles into grams using the molar mass of Mg(NO₃)₂.

(0.02856 moles Mg(NO₃)₂ )*(148grams/mole) = 4.23 grams Mg(NO₃)₂ (3 sig figs)

I hope this helps,

Bob