Question

A dioxin-contaminated water source contains 0.083% dioxin by mass. How much dioxin is present in 3.3 L of this water? Assume a density of 1.00 g/mL

Answer 1

Answer:

2.739 grams of dioxin is present in 3.3 L of this water.

Explanation:

Volume of water , V = 3.3 L = 3.3 × 1000 mL = (1 L = 1000 mL)

Mass of water = m

Density of water  =d = 1.00 g/mL

$d=\frac{m}{V}$

$m=d\times V=1.00 g/mL\times 3.3\times 1000 mL=3300 g$

Percentage of dioxon in water = 0.083%

Let the mass of dioxon present in 3300 grams of water be = x

$0.083\%=\frac{x}{3300 g}\times 100$

$x=\frac{0.083}{100}\times 3300 g=2.739 g$

2.739 grams of dioxin is present in 3.3 L of this water.