Question

A 150ml of 0.15 M NaCl solution is added to another 250 ml of 0.2M NaCl solution and the resulting solution was mixed with a 150 ml NaCl solution of unknown molarity. molarity of the final solution was formed to be 0.25M. then the unknown molarity of the solution is

Answer 1

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Answer 2

The unknown molarity of NaCl solution is 0.433 M.

Given,

Solution₁ = 150 ml of 0.15 M NaCl

Solution₂ = 250 ml of 0.20 M NaCl

Solution₃ = 150 ml of NaCl

Final molarity of solution = 0.25 M

To Find,

Molarity of Solution₃

Solution,

For mixtures of various solutions, the molarity of the final solution is given by

Final Molarity = $\frac{Sum of moles of all the solutions}{Total volume}$

Sum of moles of solution = M₁V₁ + M₂V₂ + M₃V₃ + .........

Total Volume = V₁ + V₂ + V₃ + .........

M₁ = 0.15 M

M₂ = 0.20 M

M₃ = Unknown

V₁ = 150 ml = 0.15 L

V₂ = 250 ml = 0.25 L

V₃ = 150 ml = 0.15 L

Final Molarity = M = 0.25 M

Let the unknown molarity M₃ be x

0.25 = $\frac{0.15(0.15)+0.20(0.25)+x(0.15)}{0.15+0.25+0.15}$

0.25 = $\frac{0.0225+0.05+0.15x}{0.55}$

0.25 x 0.55 = 0.0725 + 0.15x

0.1375 - 0.0725 = 0.15x

x = $\frac{0.065}{0.15}$

x = 0.433 M

Thus, the unknown molarity is 0.433 M.

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