Question

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.55 g PbCl 2 ( s ) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb ( NO 3 ) 2 ( aq ) solution.

Answer 1

Answer:

The molarity of the $Pb(NO_3)_2$ solution = 0.1895 M

Explanation:

The mass of the $PbCl_2$ obtained = 10.55 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,

$Moles= \frac{10.55\ g}{278.1\ g/mol}$

$Moles_{PbCl_2}= 0.0379\ mol$

From the reaction:-

$Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3$

1 mole of $PbCl_2$ is produced from 1 mole of $Pb(NO_3)_2$

Also,

0.0379 mole of $PbCl_2$ is produced from 0.0379 mole of $Pb(NO_3)_2$

Mole of $Pb(NO_3)_2$ = 0.0379 mol

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:

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

Thus,

$Molarity=\frac{0.0379}{0.2}\ M = 0.1895\ M$

The molarity of the $Pb(NO_3)_2$ solution = 0.1895 M