Question

A chemist needs 30mL of a 12% acid solution for an experiment. The lab has available a 10% solution and a 25% solution. How many milliliters of the 10% solution and how many milliliters of the 25% solution should the chemist mix to make the 12% solution?

Answer 1

parts of 10% soln, x

parts of 25% soln, y

total soln, x+y =30

{x(0.1) + y(0.25)}/(x + y) = 0.12...eqn 1

x + y = 30...eqn 2

from eqn 2...=》 x = 30-y

subst for x in eqn 1...

=》 {(30-y)(0.1) + y(0.25)}/ 30-y+y = 0.12

=》 (3-.1y+.25y)/30 =0.12

=》 3+.15y = 3.6

=》 .15y = .6

=》 y =4

using x = 30 - y = 26

ans

26ml of 10% soln

4ml of 25% soln

Answer 2

Using a system of equations to represent the scenario, 26mL of 10% solution and 4mL of 25% solution would be required.

Let ;

• Amount of 10% solution = a
• Amount of 25% solution = b

• a + b = 30 - - - - (1)

• 0.1a + 0.25b = (0.12 × 30)
• 0.1a + 0.25b = 3.6 - - - (2)

From (1)

• a = 30 - b - - - (3)

Substitute (3) into (2)

0.1(30 - b) + 0.25b = 3.6

3 - 0.1b + 0.25b = 3.6

0.15b = 0.6

b = 0.6/0.15

b = 4

From (3) :

a = 30 - 4

a = 26

Hence, 26mL of 10% solution and 4mL of 25% solution would be required.

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