Solving Mixture Problems Part Two

Solving Mixture Problems Part Two

This hub is a sequel to the hub “Solving Mixture Problems”. In this hub I present additional four problems with their solution. Hope you will enjoy this.

Problem Number One :

How many gallons of water must be evaporated from 100 gallons of 75 % salt solution to increase the concentration to 90 % ?

Solution :

Let X = Amount of water that must be evaporated

Original solution :

Amount of solution : 100 gallons

Amount of salt in the solution : 75 % of 100 gallons

Amount of water in the solution : 25 % of 100 gallons

Resulting solution :

Amount : 100 – X

Amount of salt : 90% of (100-X)

Amount of water 10% 0f (100-X)

Working equation:

.25(100) – X = .10 (100 – X )

25(100) – 100X = 10 (100 – X)

2500 – 100X = 1000 – 10 X

-100X + 10X = 1000 – 2500

-90 X = -1,500

90X = 1,500

(1/90) 90X = 1,500 (1/90)

X = 16.67 gallons

The amount of water that must be evaporated is about 16.67 gallons

Problem Number Two :

A radiator that holds 16 quartz is full of a solution of 30% alcohol. How much of this solution must be drawn off and replaced with pure alcohol in order that the contents of the radiator may be 55% alcohol ?

Solution :

Let X = Amount of original solution that must be drawn and replaced with pure alcohol

Original solution:

Amount of solution : 16 quartz

Amount of alcohol : 30% of 16 quarts

Amount of water : 55 % of 16 quartz

Resulting solution :

Amount of solution: 16 quatz

Amount of alcohol : 55% of 16 quartz

Amount of water : 45% of 16 quartz

Working equation :

.30(16) - .30X + X = .55(16)

30(16) - 30X +100X = 55(16)

480 + 70X = 880

70X = 880 – 480

70X/70 = 400/70

X = 5.71 quartz

Problem Number Three :

A 200 millimeter shampoo with 80% cleansing power is to be diluted with water. This is done by drawing out some amount and replacing it with water. Agnes wants a mild shampoo with 60% cleansing power. How much must she draw off and replace?

Solution:

Let X = Amount of solution must be drawn off and replace

Original solution :

Amount of solution : 200 milliliters

Amount of cleansing power : 80% of 200 ml

Amount of water : 20% of 200 ml

Resulting Solution :

Amount of solution : 200 ml

Amount of cleansing power : 60% of 200 ml

Amount of water : 40% of 200 ml

Working equation :

.20(200) - .20X + X = .40 (200)

20 (200) – 20X + 100 X = 40 (200)

4000 + 80X = 8000

80X = 8000 – 4000

80X = 4000

80X/80 = 4000/80

X = 50 milliliters

Problem Number Four :

Lynn, a chemist mixed 40 mL of 8% HCl acid with 60 mL of 12% HCl acid solution. She used a portion of this solution and replaced it with distilled water. If the new solution tested 5.2 % HCl acid, how much of the original solution did she use ?

Let X = Amount of original solution used and replaced with distilled water.

Step One : Find first the percentage HCl in the original mixture

.08(40) + .12(60) = X 100

8(40) + 12(60) = 10000X

320 + 720 = 10,000X

1,040 = 10,000X

10,000X = 1,040

10,000X/10,000 = 1,040 /10,000

X = 10. 4% HCl in the original mixture

Step two :

Amount of original solution: 100mL

Amount of HCl : 10.4% of 100 mL

Amount of Water : 89.6 % of 100 mL

Resulting solution :

Amount : 100 mL

Amount HCl : 5.2 % of 100 mL

Amount of water : 94.8% of 100 mL

Working equation :

.896(100) - .896X + X = .948(100)

896(100) – 896 X + 1000X = 948(100)

89,600 + 104X = 94,800

104X = 94,800 – 89,600

104X = 5,200

104X/104 = 5,200/104

X = 50 milliliters of the original mixture must be drawn and replaced with distilled water.