Chemistry, 26.03.2020 21:08 mangowammy

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

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Answer from: naruto63

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 of KOH =?

Mole = mass / volume

Mole = 10 / 56

Mole of KOH = 0.179 mole

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 =?

Concentration = mole / Volume

Concentration = 0.179 / 0.15

Concentration of stock solution = 1.19 M

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

Concentration of diluted solution (C₂) =? C₁V₁ = C₂V₂

1.19 × 15 = C₂ × 65

17.85 = C₂ × 65

Divide both side by 65

C₂ = 17.85 / 65

C₂ = 0.27 M

Therefore, the concentration of the final solution is 0.27 M

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Answer from: oscardiazbet8803

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.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

V_s = volume of solution in ml = 150 ml

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles$

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

$Molality=\frac{0.178\times 1000}{150}=1.19M$

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

$1.19\times 15.0=M_2\times 65.0$

M_1=0.275M

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

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A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of s...

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