Question

A solution of NaCl (aq) is added slowly to a solution of lead nitrate, Pb(NO3)2 ( aq ), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.91 g PbCl2 (s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2 (aq) solution.

Answer 1

Answer:

0.19615 M

Explanation:

Calculation of the moles of $PbCl_2$ as:-

Mass = 10.91 g

Molar mass of $PbCl_2$ = 278.1 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.91\ g}{278.1\ g/mol}$

$Moles= 0.03923\ mol$

According to the reaction shown below:-

$Pb(NO_3)_2_{(aq)} + 2NaCl_{(aq)}\rightarrow PbCl_2_{(s)} + 2NaNO_3_{(aq)}$

1 mole of $PbCL_2$ is produced when 1 mole of $Pb(NO_3)_2$ undergoes reaction.

0.03923 mole of $PbCL_2$ is produced when 0.03923 mole of $Pb(NO_3)_2$ undergoes reaction.

Mole of $Pb(NO_3)_2$ = 0.03923 mole

Volume = 200.0 mL = 0.2 L ( 1 mL = 0.001 L )

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,

$Molarity=\frac{0.03923}{0.2}\ M= 0.19615\ M$

0.19615 M is the molarity of the $Pb(NO_3)_2$ solution.