Benzene is a carcinogenic (cancer-causing) compound. A benzene-contaminated water sample contains 3.1×10−5% benzene by mass.
What volume of the water in liters contains 225mg of benzene? (Assume that the density of the solution is 1.0 g/mL.)
(g solute/g solution)*100 = 3.1E-5
(0.225/g soln)*100 = 3.1E-5
Solve for g solution (in mL) and convert to L.
To find the volume of water containing 225mg of benzene, we first need to calculate the mass of the water.
Given that the benzene-contaminated water sample contains 3.1×10−5% benzene by mass, we can assume that 100% - 3.1×10−5% = 99.99969% of the solution is water.
Let's assume we have 1 gram of the solution.
Therefore, the mass of benzene in 1 gram of the solution would be (3.1×10−5 / 100) × 1g = 3.1×10−7 g.
Since the density of the solution is given as 1.0 g/mL, the mass of the water in 1 mL (or 1 cm³) would be 1 g.
Therefore, the mass of water in 1 gram of the solution would be 1 g - 3.1×10−7 g = 0.99999969 g.
Now we can calculate the volume of water in 1 gram of the solution.
Since the density of water is 1.0 g/mL, the volume of water in 1 gram of the solution would be 0.99999969 mL.
To convert mL to liters, we divide by 1000.
So, the volume of water in 1 gram of the solution in liters would be 0.99999969 mL / 1000 = 9.9999969×10^-4 L.
Now that we know the volume of water in 1 gram of the solution, we can use this to find the volume of water containing 225mg of benzene.
Given that we have 225mg of benzene, let's calculate the mass of the water needed.
Mass of water = Mass of benzene / % of water in 1 gram of the solution
Mass of water = 225mg / 0.99999969
Converting the mass of water to liters, we divide by 1000.
Volume of water = Mass of water / density of water
Volume of water = (225mg / 0.99999969) / 1000 = 0.225 L
Therefore, the volume of water required to contain 225mg of benzene is 0.225 liters.
To find the volume of water that contains 225mg of benzene, we can use the given mass percentage of benzene in the water sample.
First, let's calculate the mass of benzene present in the water sample.
Given:
Mass percentage of benzene in water = 3.1×10^(-5)%
Mass of the water sample = 225mg
To determine the mass of benzene in the water sample, we will use the mass percentage formula:
Mass of benzene = (Mass percentage of benzene/100) × Mass of the water sample
Mass of benzene = (3.1×10^(-5)/100) × 225mg
Mass of benzene = (3.1×10^(-5) × 225mg) / 100
Mass of benzene = 0.06975mg
Now, let's convert the mass of benzene to volume using the density of the solution.
Given:
Density of the solution = 1.0 g/mL
Mass of benzene = 0.06975mg
To determine the volume of the water, we will use the density formula:
Density = Mass/Volume
Volume = Mass/Density
Volume = 0.06975mg / 1.0 g/mL
As the units do not match, let's convert milligrams (mg) to grams (g) first.
1 mg = 0.001 g
Volume = 0.06975g / 1.0 g/mL
Volume = 0.06975 mL
Finally, let's convert milliliters (mL) to liters (L).
1 L = 1000 mL
Volume = 0.06975 mL / 1000
Volume = 0.00006975 L
Therefore, the volume of water containing 225mg of benzene is 0.00006975 liters.