Question

A 3.00 L sample of paint that has a density of 4.65 g/mL is found to contain 33.1 g lead (II) nitride. what is mass percentage of lead in the paint and what is the ppm?

Answer 1

Density relates the mass of a compound to its volume. We will first find the mass of the solution using the density and volume given. We have the following equation:

`\rho=\text{ density = }(Mass)/(Volume)`

We clear the mass,

`\begin{gathered} Mass\text{ = }\rho* Volume \\ Mass\text{ =}4.65(g)/(mL)*3000mL \\ Mass\text{ =}13950\text{ g} \end{gathered}`

So, the mass of paint is 13950 g. Now the mass percentage of the lead in the paint will be calculated as follows:

`\begin{gathered} \%mass=\frac{\text{ Mass of lead (II) nitride}}{Mass\text{ of paint}}*100 \\ \%mass=\frac{33.1\text{ g}}{13950\text{ g}}*100 \\ \%mass=0.24\% \end{gathered}`

The mass percentage of lead in the paint is 0.24%.

ppm concentration means the quantity of solute in milligrams (mg) is in a liter (L) of a solution. So, to calculate ppm concentration we will divide the milligrams of lead (II) nitride between the liters of paint:

`\begin{gathered} \text{ppm = }\frac{\text{33.1g of lead}*(1000mg)/(1g)}{3L} \\ \text{ppm =11033ppm of lead} \end{gathered}`

The ppm of lead in the paint is 11033ppm.