Question

You have a stock solution of 14.0 M NH3. How many milliliters of this solution should you dilute to make 1250 mL of 0.300M NH3?

Answer 1

First take 1250 mL of 0.3 M NH3

Molarity = number of moles of solute / volume of the solution in L

0.3 M = number of moles of solute / 1.25 L

Number of moles of solute = 1.25 * 0.3 = 0.375 mol of NH3 are required

Now We have the source at 14.0 M NH3 and we can determine how many Litres/millilitres of this concentration can provide us the 0.375 mol which is required for making 1250 mL of 0.300 M NH3

Molarity = number of moles of solute / Volume of the solution litres

14.0 M = 0.375 mol/ Volume of the solution in L

Volume of the solution in L = 0.375 mol / 14.0 M = 0.02678 L

So millilitres of the 14.0 M NH3 required to prepare 1250 mL of 0.300 M NH3 = 26.78 mL