Question

Paul has an 50​% methyl alcohol solution. He wishes to make a gallon of a solution by mixing his methyl alcohol solution with water. If 128 ​ounces, or a​ gallon, of solution should contain 5​% methyl​ alcohol, how much of the 50​% methyl alcohol solution and how much water must be​ mixed?

Answer 1

To make one gallon of 5% methyl alcohol we require 115.2oz of water and 12.8oz of 50% methyl alcohol solution

Step-by-step explanation:

Required solution = 1 gallon of 5% methyl alcohol or1 gallon 5/100 moles per dm3 of methyl alcohol

Number of moles in required solution = 0.05moles/dm3 × 1gallon

Where 1 gallon = 0.00378dm3

= 0.05moles/dm3 × 0.00378dm3 = 1.892×10-4 moles

The quantity of the 50% solution with Paul required is given by

50% = 50/100 = 0.5 moles/dm3

Volume of 50% solution of methyl alcohol that contains 1.892×10-4 moles

= concentration = (Number of moles)/(volume of dm3) of solution thus

Volume of solution = (Number of moles)/(concentration)

1.892/0.5 =3.785×10-4dm3 of the 50% solution is required to make 1 gallon of 5% solution

3.785×10-4dm3 = 12.8oz

Thus to make 1 gallon or 128oz solution of 5% alcohol requires

128oz - 12.8 oz = 115.2oz of water and 12.8oz of 50% methyl alcohol solution