Mass percent related problems

[apiwowar]

a pb-contaminated water sample contains .0011% pb by mass. how much of the water (in mL) contains 150mg of pb. the density of the solution is 1.0g/mL.

to solve that would you just start with 150mg of pb then convert that to grams, use the mass % as a conversion factor by making it 100g sample/ .0011g pb then use the remaining mass as the mass of the water so i end up with grams of water then use the density to figure out milliLiters of water?

150mg pb * 1gpb/10^3mgpb * 100g sample/.0011g Pb * 99.9989 g water/ 100g sample * 1ml / 1.0g = 13636.21364 mL

did i do that right or is that wrong?

another problem

how many moles of potassium sulfate are present in a 240.0 g sample of a 35% potassium sulfate sample by mass?

this is my calculation:

240.0g sample * 35g k2so4/100g sample * 1mole k2so4/ 174.26g k2so4 = .4766 moles k2so4

does that look right or did i do it wrong?

Homework Statement

 

[nate808]

A water sample contains 0.0011% lead (Pb) by mass and has a density of 1.0 g/mL. How many milliliters of water contain 150 mg of Pb?

Homework Equations

mass % = (mass of solute / mass of solution) * 100

density = mass / volume

The Attempt at a Solution

To solve this problem, we first need to convert the mass of Pb from milligrams (mg) to grams (g).

150 mg Pb * 1 g Pb / 1000 mg Pb = 0.150 g Pb

Next, we can use the mass % as a conversion factor to find the mass of the entire sample.

0.150 g Pb * (100 g sample / 0.0011 g Pb) = 13636.36 g sample

Since we are looking for the volume of water, we can subtract the mass of Pb from the total sample mass to find the mass of water.

13636.36 g sample - 0.150 g Pb = 13636.21 g water

Finally, we can use the density equation to find the volume of water in milliliters.

13636.21 g water * (1 mL / 1.0 g) = 13636.21 mL water

Therefore, 13636.21 mL of water contains 150 mg of Pb.

For the second problem, we can use a similar approach.

First, we convert the mass of the sample from grams to kilograms.

240.0 g sample * 1 kg / 1000 g = 0.240 kg sample

Next, we can use the mass % as a conversion factor to find the mass of potassium sulfate (K2SO4) in the sample.

0.240 kg sample * (35 g K2SO4 / 100 g sample) = 0.084 kg K2SO4

Now, we can use the molar mass of K2SO4 (174.26 g/mol) to convert the mass of K2SO4 to moles.

0.084 kg K2SO4 * (1000 g / 1 kg) * (1 mol K2SO4 / 174.26 g K2SO4) = 0.4816 mol K2SO4

Therefore, there are 0.4816 moles of K2SO4 present in a 240.0