How would you prepare 300 mL of 0.750 M KCl?

1 Answer
Feb 24, 2018

16.7 grams of KCl diluted to a final volume of 300mL.

Explanation:

Molarity is equal to the number of moles of a solute per 1 liter of solution. (It's a way to measure concentration in chemistry).

In this case, the KCl is the solute, and 0.750 M tells you how many moles of it we have per 1 liter of solution.

Convert moles of KCl to grams using its molar mass, 74.5513 g/mol.

#0.750 mol * ((74.5513g)/(1 mol)) # = 55.883 grams.

In one liter of solution, there are 55.883 grams of KCl.

But the question wants to know about 300mL (which is equal to 0.3L)

#0.3L * (55.883g)/(1L) # = 16.7499 grams, round to 3 SF = 16.7 grams.

In a lab setting, if you were to actually prepare this mixture, you would first measure out 16.7 grams of your KCl, and add this to a volumetric flask. Then you would fill with water to the line on the neck of the bottle (300mL mark in this example.) Always do it in this order, otherwise you will not have an accurate volume of solution.