# Question 3084b

Dec 16, 2015

$\text{123 g}$

#### Explanation:

The trick here is to make sure that you account for the water of hydration when weighing out the mass of magnesium sulfate heptahydrate needed to produce that target magnesium sulfate solution.

The first thing to do here is figure out how many moles of magnesium sulfate, ${\text{MgSO}}_{4}$, are needed to make a $\text{0.200-M}$, $\text{2500.0 mL}$ solution.

As you know, molarity is defined as moles of solute per liter of solution. In your case, the target solution would contain

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{M g S {O}_{4}} = {\text{0.200 M" * 2500.0 * 10^(-3)"L" = "0.500 moles MgSO}}_{4}$

Use magnesium sulfate's molar mass to determine how many grams of the compound would contain this many moles

0.500 color(red)(cancel(color(black)("moles"))) * "120.30 g"/(1color(red)(cancel(color(black)("mole")))) = "60.15 g MgSO"_4

The next thing to do here is calculate the percent composition of magnesium sulfate heptahydrate, $\text{MgSO"_4 * 7 "H"_2"O}$.

To do that, use the fact that one mole of hydrate contains

• one mole of magnesium sulfate, ${\text{MgSO}}_{4}$
• seven moles of water, $\text{H"_2"O}$

The percent be mass of magnesium sulfate in one mole of magnesium sulfate heptahydrate will thus be

(120.30 color(red)(cancel(color(black)("g"))))/(120.30 + 7 * 18.02 color(red)(cancel(color(black)("g")))) xx 100 = 48.815%

This means that every $\text{100 g}$ of hydrate will contain $\text{48.815 g}$ of magnesium sulfate.

You need to provide $\text{60.15 g}$ of magnesium sulfate to the solution, so the mass of hydrate that contains this many grams of magnesium sulfate will be

60.15 color(red)(cancel(color(black)("g MgSO"_4))) * "100 g hydrate"/(48.815color(red)(cancel(color(black)("g MgSO"_4)))) = "123.22 g"

Rounded to three sig figs, the number of sig figs you have for the molarity of the target solution, the naswer will be

m_"hydrate" = color(green)("123 g MgSO"_4 * 7"H"_2"O")#