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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very small mass. A solution was prepared by dissolving 0.360 g of KNO3 in enough water to make 500. mL of solution. A 10.0 mL sample of this solution was transferred to a 500.0-mL volumetric flask and diluted to the mark with water. Then 10.0 mL of the diluted solution was transferred to a 250.0-mL flask and diluted to the mark with water. What is the final concentration of the KNO3 solution?

Respuesta :

Answer: [tex]5.70\times 10^{-6}M[/tex]

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

Given : 0.360 g of [tex]KNO_3[/tex] is dissolved in 500 ml of solution.

[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.360g}{101g/mol}=3.56\times 10^{-3}mole[/tex]  

[tex]V_s[/tex] = volume of solution  = 500 ml

[tex]Molarity=\frac{3.56\times 10^{-3}\times 1000}{500}=7.12\times 10^{-3}M[/tex]

According to the neutralization law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock solution = [tex]7.12\times 10^{-3}M[/tex]

[tex]V_1[/tex] = volume of stock solution = 10.0 ml

[tex]M_2[/tex] = molarity of diluted solution = ?

[tex]V_2[/tex] = volume of diluted solution = 500.0 ml

[tex]7.12\times 10^{-3}M\times 10.0=M_2\times 500.0[/tex]

[tex]M_2=1.42\times 10^{-4}M[/tex]

b)  On further dilution

[tex]M_1[/tex] = molarity of stock solution = [tex]1.42\times 10^{-4}M[/tex]

[tex]V_1[/tex] = volume of stock solution = 10.0 ml

[tex]M_2[/tex] = molarity of diluted solution = ?

[tex]V_2[/tex] = volume of diluted solution = 250.0 ml

[tex]1.42\times 10^{-4}M\times 10.0=M_2\times 250.0[/tex]

[tex]M_2=5.70\times 10^{-6}M[/tex]

Thus the final concentration of the [tex]KNO_3[/tex] solution is [tex]5.70\times 10^{-6}M[/tex]

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