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

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

$V_s$ = volume of solution in ml

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

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

$Molarity=\frac{0.00031mol\times 1000}{100ml}=0.0031M$

According to the dilution law,

$(stock)M_1V_1=M_2V_2(dilute)$

Putting in the values we get:

$0.0031\times 5.00=M_2\times 50.00$

$M_2=0.00031M$

Therefore, the molarity of diluted solution is 0.00031 M