A researcher has saline solutions with concentrations of 8% and 17%. He wants to mix some of each solution to get 30 milliliters

Question

2 years ago

6

of solution with an 11% concentration. How many milliliters of each solution does he need to mix together?

1 answer:

VladimirAG [237] · Answer 1 · 2023-05-12T05:06:18+03:00

The researcher needs to add 20 ml of 8% solution and 10 ml of 17% solution to make a 11% solution

<h3>What do you mean by Concentration ?</h3>

Concentration is the measure of the mass of solute in certain volume of solution.

The parameters useful in measuring concentration is Molarity , Normality , Molality etc.

It is given that concentration of saline solutions with concentrations of 8% and 17%.

A mixture of 11% 30 ml concentration solution has to be made

The formula for mixing mixtures is given by

$\rm M = \dfrac{m_1 x_1 +m_2x_2}{x_1+x_2}$

Here m₁ and m₂ represents the concentration and x₁ and x₂ represents the volume of solution added x₁ = x and x₂ 30-x

M = ( 8 * x +(17*(30-x))/(30)

30*11 = 8x +510 -17x

330 = -9x +510

9x = 180

x =20ml

Therefore the researcher needs to add 20 ml of 8% solution and 10 ml of 17% solution to make a 11% solution.

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