if the average concentration of PCBs in the body tissue of a human is 4.0 ppm, what mass of the PCB is present in a 64 kg person? is this a significant amount in regards to the health of the person?
What is the concentration of a solution of sodium nitrate that contains 35.0 g of the solute in a 1.5 L solution?
A typical household ammonia solution has a concentration o 1.24 M. What volume of this solution would contain .500 mol of NH3? What volume of this solution would contain 10.0 g of NH3?
What volume of concentrated 17.8 M sulfuric acid would a laboratory technician need to make 2.00 L of a .200 mol/L solution by dilution of the original, concentrated solution?
See your duplicate above.
To answer these questions, we will use the formula:
Mass of substance = Concentration x Volume
1. Mass of PCB in a 64 kg person:
The average concentration of PCBs in the body tissue of a human is given as 4.0 ppm. Since 1 ppm is equal to 1 mg/kg, the concentration can be converted to mg/kg. So, 4.0 ppm is equal to 4.0 mg/kg.
Given the mass of the person as 64 kg, we can calculate the mass of PCBs:
Mass of PCBs = Concentration x Mass of the person
Mass of PCBs = 4.0 mg/kg x 64 kg
To find the mass in grams:
Mass of PCBs = (4.0 mg/kg x 64 kg) / 1000
Mass of PCBs = 0.256 g
So, there would be 0.256 grams of PCBs present in a 64 kg person.
Regarding the significance in terms of the health of the person, it depends on the toxicity and accepted exposure limits of PCBs. You would need to refer to relevant guidelines or consult a healthcare professional to determine if this amount is significant or not.
2. Concentration of sodium nitrate solution:
Given the mass of the solute (sodium nitrate) as 35.0 g and the volume of the solution as 1.5 L, we can use the formula:
Concentration = Mass of solute / Volume of solution
Concentration = 35.0 g / 1.5 L
Simplifying the equation, we get:
Concentration = 23.3 g/L
So, the concentration of the sodium nitrate solution is 23.3 g/L.
3. Volume of 1.24 M ammonia solution containing 0.500 mol of NH3:
To find the volume, we can rearrange the formula:
Volume = Number of moles / Concentration
Volume = 0.500 mol / 1.24 M
Volume = 0.403 L or 403 mL
So, the volume required is 0.403 liters or 403 milliliters.
Volume of 1.24 M ammonia solution containing 10.0 g of NH3:
First, we need to convert the mass of ammonia to moles using the molar mass of NH3 (17.031 g/mol):
Number of moles = Mass of NH3 / Molar mass of NH3
Number of moles = 10.0 g / 17.031 g/mol
Number of moles = 0.587 mol
Now we can use the same formula as above:
Volume = Number of moles / Concentration
Volume = 0.587 mol / 1.24 M
Volume ≈ 0.473 L or 473 mL
So, the volume required is approximately 0.473 liters or 473 milliliters.
4. Volume of concentrated 17.8 M sulfuric acid for 2.00 L of 0.200 mol/L solution:
To dilute the concentrated solution, we can use the formula:
Concentration1 x Volume1 = Concentration2 x Volume2
Given:
Concentration1 = 17.8 M
Volume1 = ?
Concentration2 = 0.200 mol/L
Volume2 = 2.00 L
Rearranging the formula to find Volume1:
Volume1 = (Concentration2 x Volume2) / Concentration1
Volume1 = (0.200 mol/L x 2.00 L) / 17.8 M
Volume1 ≈ 0.022 L or 22 mL
So, the laboratory technician would need approximately 0.022 liters or 22 milliliters of concentrated 17.8 M sulfuric acid to make a 2.00 L solution with a concentration of 0.200 mol/L by dilution.