Question

A stock solution was created by adding 0.1042 g of lead (II) nitrate to a 100.00 mL volumetric flask and diluting to volume with deionized water. A diluted solution was then created by removing 5.00 mL of the stock solution and placing into into a 50.00 mL volumetric flask and then diluting to volume with deionized water. What is the concentration (in molarity, M) of the diluted solution

Answer 1

Answer: The molarity of diluted solution is 0.00031 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

`V_s` = volume of solution in ml

moles of
`Pb(NO_3)_2` =
`\frac{\text {given mass}}{\text {Molar mass}}=(0.1042g)/(331.2g/mol)=0.00031mol`

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

`Molarity=(0.00031mol* 1000)/(100ml)=0.0031M`

According to the dilution law,

`(stock)M_1V_1=M_2V_2(dilute)`

Putting in the values we get:

`0.0031* 5.00=M_2* 50.00`

`M_2=0.00031M`

Therefore, the molarity of diluted solution is 0.00031 M