Answer: 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.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml = 150 ml
moles of solute =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molality=\frac{0.178\times 1000}{150}=1.19M[/tex]
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = molarity of stock solution = 1.19 M
[tex]V_1[/tex] = volume of stock solution = 15.0 ml
[tex]C_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 65.0 ml
Putting in the values we get:
[tex]1.19\times 15.0=M_2\times 65.0[/tex]
[tex]M_1=0.275M[/tex]
Therefore, the concentration of KOH for the final solution is 0.275 M
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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