What is the molarity of a solution of ammonium chloride prepared by diluting 50.00 mL of a 3.79 M #NH_4Cl# solution to 2.00 L?

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Answer 1 · 2016-07-28T14:22:54

0.09475 moles per litre.

How many moles of solute did you start with?

Use #M = n/V#

Where M is molarity in moles/litre, n is number of moles, and V is number of litres.

Rearrange for n: #n = M.V# = 3.79 x 0.05 = 0.1895 moles.

After diluting, how many moles of solute will you have? Answer: The same. But now the solution volume will be 2 litres.

Therefore: #M = n/V# = 0.1895 / 2 = 0.09475 mol/litre.

And if you want to make it even simpler - you are diluting 50 ml to 2000 ml, which is a dilution of 40 times.

So simply divide your starting molarity by 40. 3.79 divided by 40 is 0.09475.