Question

A chemist is mixing two solutions, solution A and solution B. Solution A is 15% water and solution Bis 20% water. She already has a

beaker with 10mL of solution A in it. How many mL of solution B must be added to the beaker in order to create a mixture that is 18%
water?

Answer 1

15 ml

Explanation:

We are told

Solution A = 15% of water

Solution B = 20% of water

Let's assume, the entire solution = 100ml

We are told that in the beaker we have 10 ml of Solution A already

Mathematically,

100 ml = 15%

10 ml = X

100ml × X = 15 × 10

X = 150/ 100

X = 1.5%

Hence in the beaker, we have 1.5% of water from Solution A

We are asked to find how many ml of solution B must be added to make the solution have 18% of water

Let y = number of ml of solution B

Hence

10 ml × 15%(0.15) = 1.5 ml of water - Equation 1

y ml × 20%( 0.20) = 0.20y ml of water ...... Equation 2

Add up the above equation

10ml + y ml ×18% (0.18) = 1.5 + 0.20y

(10 + y)(0.18) = 1.5 + 0.20y

1.8 + 0.18y = 1.5 + 0.20y

Collect like terms

1.8 - 1.5 = 0.20y - 0.18y

0.3 = 0.02y

y = 0.3/0.02

y = 15ml

Therefore,15mL of solution B must be added to the beaker in order to create a mixture that is 18% water