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
[tex]d=\frac{m}{V}[/tex]
[tex]m=d\times V=1.00 g/mL\times 3.3\times 1000 mL=3300 g[/tex]
Percentage of dioxon in water = 0.083%
Let the mass of dioxon present in 3300 grams of water be = x
[tex]0.083\%=\frac{x}{3300 g}\times 100[/tex]
[tex]x=\frac{0.083}{100}\times 3300 g=2.739 g[/tex]
2.739 grams of dioxin is present in 3.3 L of this water.
