Question

A chemist has two solutions: a 50% methane solution and an 80% methane solution. He wants 100 mL of a 70% methane solution. How many mL of each solution does he need to mix?

Answer 1

x = 33.3 mL 50 % solution

y = 66.6 mL 80% solution

Explanation:

given data

methane solution = 50%

methane solution = 80 %

want = 100 mL of a 70% methane

to find out

how many mL of each solution does he need to mix

solution

we consider x mL of 50 % solution

y mL of 80 % solution

so x + y = 100 .........................1

multiply 0.5 to equation 1

0.5 x + 0.5 y = 0.5 ( 100 ) .....................2

and 0.5 x + 0.8 y = 0.7 ( 100) .....................3

so from equation 2 and 3

y = 66.6

so x = 33.3

so

x = 33.3 mL 50 % solution

y = 66.6 mL 80% solution

Answer 2

33.33 mL of 50%

66.67 of 80%

Explanation:

Let x = quantity of 50% methane.

Then,

0.50x + 0.80(100 - x) = 100 × 0.70

⇒ 0.50 x + 80 - 0.80 x = 70

⇒ -0.30 x = 70 - 80

⇒ -0.30 x = -10

⇒
`x=(10)/(0.3)`

⇒ x = 33.33 mL of 50%

and, 100 - 33.33 = 66.67 of 80%