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* 10^(-6)M`

Step-by-step explanation:

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

`Molarity=(n* 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}}=(0.360g)/(101g/mol)=3.56* 10^(-3)mole`

`V_s` = volume of solution = 500 ml

`Molarity=(3.56* 10^(-3)* 1000)/(500)=7.12* 10^(-3)M`

According to the neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = molarity of stock solution =
`7.12* 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* 10^(-3)M* 10.0=M_2* 500.0`

`M_2=1.42* 10^(-4)M`

b) On further dilution

`M_1` = molarity of stock solution =
`1.42* 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* 10^(-4)M* 10.0=M_2* 250.0`

`M_2=5.70* 10^(-6)M`

Thus the final concentration of the
`KNO_3` solution is
`5.70* 10^(-6)M`