...A dioxin-contaminated water source contains 0.044% dioxin by mass. How much dioxin is present in 6.5 L of this water? (Assume that the density of the solution is 1.00 g/mL.)
thanks
6500grms*.00044= XXXX grams
Hint it is between 1 and 6 grams
0.044% means 0.044 g/100 g soln or
0.044 g/100 mL (Technically this is 0.044 g/(100-0.044) but for very dilute solutions we usually ignore that small difference.
0.044 g/100 is how many g in 6.5L? That is
0.044 g x 6,500 mL/100 mL = ??
so would it be 2.86?
yes, right.
thanks:)
To calculate the amount of dioxin present in 6.5 L of water, you'll first need to determine the mass of the solution. Since the density of the solution is given as 1.00 g/mL, we can convert the volume of water from liters to grams.
The density tells us that 1 mL of the solution weighs 1.00 g. So, to find the mass of 6.5 L of the solution, we'll convert liters to milliliters:
1 L = 1000 mL
Therefore, 6.5 L x 1000 mL/L = 6500 mL
Now that we have the volume of the solution in milliliters, we can convert it to grams by multiplying it by the density:
6500 mL x 1.00 g/mL = 6500 g
Next, we'll calculate the mass of dioxin present in the solution. The concentration of dioxin is given as 0.044% by mass. This means that for every 100 grams of the solution, there are 0.044 grams of dioxin.
To find the mass of dioxin in 6500 g of solution, we can set up a proportion:
0.044 g/100 g = x g/6500 g
Cross-multiplying, we get:
100 * x = 0.044 * 6500
x = (0.044 * 6500) / 100
x ≈ 2.86 g
Therefore, there is approximately 2.86 grams of dioxin present in 6.5 L of the contaminated water source.