Question

How many grams of MgCl2 would be needed to make 2.3 liters of a 0.33 molar solution? Round your answer to the nearest 0.01, and remember to include both units (properly abbreviated) and substance.

Answer 1

Using the following equation:

`M=(n)/(L)`

Where M represents the concentration (Molarity), n represents the number of moles of solute and L represents the liters of solution.

If we replace the values of the problem:

`0.33M=(n)/(2.3L)\to n=0.33M\cdot2.3L\to n=0.759molesMgCl_2`

To find the number of grams, what we do is to multiply by the molar mass of MgCl2, which is 95.211g:

`0.759molesMgCl_2\cdot\frac{^{}95.211gMgCl_2}{1molesMgCl_2}=72.26gMgCl_2`

Therefore, 72.26gMgCl2 are needed.