Question

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?

Answer 1

Answer: $5.70\times 10^{-6}M$

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

Given : 0.360 g of $KNO_3$ is dissolved in 500 ml of solution.

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

$V_s$ = volume of solution  = 500 ml

$Molarity=\frac{3.56\times 10^{-3}\times 1000}{500}=7.12\times 10^{-3}M$

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $7.12\times 10^{-3}M$

$V_1$ = volume of stock solution = 10.0 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 500.0 ml

$7.12\times 10^{-3}M\times 10.0=M_2\times 500.0$

$M_2=1.42\times 10^{-4}M$

b)  On further dilution

$M_1$ = molarity of stock solution = $1.42\times 10^{-4}M$

$V_1$ = volume of stock solution = 10.0 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 250.0 ml

$1.42\times 10^{-4}M\times 10.0=M_2\times 250.0$

$M_2=5.70\times 10^{-6}M$

Thus the final concentration of the $KNO_3$ solution is $5.70\times 10^{-6}M$