# What mass of MgSO_4 . 7H_2O is required to prepare 300 mL of a 0.61 M MgSO_4 solution? Answer in...

## Question:

What mass of {eq}\rm MgSO_4 \cdot 7H_2O {/eq} is required to prepare {eq}300\ mL {/eq} of a {eq}\rm 0.61\ M\ MgSO_4 {/eq} solution? Answer in units of g.

## Molar Mass:

Molar mass is a quantity describing the total mass per given mole of a substance. The mole is known to be a unit of measurement which corresponds to {eq}\displaystyle 6.022\times 10^{23} {/eq} number of atoms or molecules of a substance.

Determine the mass of the said reagent by first finding the number of moles required for the given solution, and then we multiply it to the ratio of the moles of contained {eq}\displaystyle MgSO_4 {/eq} per molecule of {eq}\displaystyle MgSO_4 \cdot 7H_2O {/eq} and then we multiply the molar mass of {eq}\displaystyle MgSO_4 \cdot 7H_2O {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle 300\ mL\times \frac{1}{1000}\ \rm{L/mL}\times 0.61\ M\ MgSO_4\times \frac{1\ mol\ MgSO_4\cdot 7H_2O}{1\ mol\ MgSO_4}\times 246.47\ \rm{g/mol} =45\ g\ MgSO_4 \cdot 7H_2O \end{align} {/eq}