Respuesta :
Answer:
[tex]4.71\times 10^{-5} g[/tex]of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.
Explanation:
Mass of sodium arsenate =x
Mass of water = M
Density of water = d = 1000 g/L
Volume of the water , V= 1.70 L
M = [tex]d\times V= 1000 g/L\times 1.70 L=1,700 g=1.700 kg[/tex]
1 Parts per billion = 1 μg/kg = [tex]\frac{10^{-9} kg}{1 kg}[/tex]
[tex]10 ppb = \frac{x}{1.700 kg}[/tex]
[tex]x=10\times 10^{-9} \times 1.700 kg=1.700\times 10^{-8} kg[/tex]
[tex]x=1.700\times 10^{-8} kg=1.700\times 10^{-8}\times 10^3 g=1.700\times 10^{-5}g[/tex]
(1 kg =1000 g)
Moles of arsenic =[tex]\frac{1.700\times 10^{-5}g}{75 g/mol}=2.267\times 10^{-7} mol[/tex]
1 mole of sodium arsenate has 1 mol of arsenic atom.Then [tex]2.267\times 10^{-7} mol[/tex] of arsenic will found in:
[tex]1\times 2.267\times 10^{-7} mol=2.267\times 10^{-7} mol[/tex] of sodium arsenate.
Mass of [tex]2.267\times 10^{-7} mol[/tex] of sodium arsenate:
[tex]208 g/mol\times 2.267\times 10^{-7} mo=4.71\times 10^{-5} g[/tex]
[tex]4.71\times 10^{-5} g[/tex]of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.
