A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution. What is the concentration of KOH for the final solution

The concentration of the final solution is 0.27 M

We'll begin by writing calculating the mole of 10 g of KOH.

Mass of KOH = 10 g

Molar mass of KOH = 39 + 16 + 1

= 56 g/mol

Mole = mass / volume

Mole = 10 / 56

Next, we shall determine the concentration of the stock solution. This can be obtained as follow:

Mole of KOH = 0.179 mole

Volume = 150 mL = 150 / 1000 = 0.15 L

Concentration = mole / Volume

Concentration = 0.179 / 0.15

Finally, we shall determine the concentration of the diluted solution (i.e final concentration).

Concentration of stock solution (C₁) = 1.19 M

Volume of stock solution (V₁) = 15 mL

Volume of diluted solution (V₂) = 65 mL

1.19 × 15 = C₂ × 65

17.85 = C₂ × 65

Divide both side by 65

C₂ = 17.85 / 65

Therefore, the concentration of the final solution is 0.27 M

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The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,

n = moles of solute

  = volume of solution in ml = 150 ml

moles of solute =

Now put all the given values in the formula of molality, we get

According to the dilution law,

where,

= molarity of stock solution = 1.19 M

= volume of stock solution = 15.0 ml

= molarity of diluted solution = ?

= volume of diluted solution = 65.0 ml

Putting in the values we get:

Therefore, the concentration of KOH for the final solution is 0.275 M

i think its 100.

hope this !