Calculate the number of moles and the mass of the solute in each of the following solutions:(a) 2.00 L of 18.5 M H2SO4, concentrated sulphuric acid
(b)100.0 mL of 3.8 x 10-5 MNACN, the minimum Lethal concentration of sodium cyanide in blood serum
(c)5.50 L of 13.3 M H2CO, the formaldehyde used to "fix" tissue samples.
(d)325 mL of 1.8 x 10-6 M FeSO4, the minimum concentration of iron sulphate detectable by taste in drinking water.

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Tringa0 Tringa0 · Answer 1 · 2019-09-05T12:41:26+02:00

Explanation:

[tex]c=\frac{n}{V}[/tex]

c = Concentration of the solution

n = Moles of compound in solution

V = Volume of the solution

a) 2.00 L of 18.5 M of concentrated sulfuric acid.

n= ? c = 18.5 M, V = 2.00 L

[tex]18.5 M=\frac{n}{2.00 L}[/tex]

n = 37 moles of sulfuric acid

b) 100.0 mL of [tex]3.8\times 10^{-5} M[/tex] of sodium cyanide

n= ? ,c = [tex]3.8\times 10^{-5} M[/tex], V = 100.0 mL = 0.1 L

[tex]3.8\times 10^{-5} M=\frac{n}{0.1 L}[/tex]

n = [tex]3.8\times 10^{-6} moles [/tex] of sodium cyanide

c) 5.50 L of 13.3 M of concentrated formaldehyde.

n= ? c = 13.3 M, V = 5.50 L

[tex]13.3 M=\frac{n}{5.50 L}[/tex]

n = 73.15 moles of formaldehyde.

d)325 mL of [tex]1.8\times 10^{-6} M[/tex] of iron sulphate

n= ? ,c = [tex]1.8\times 10^{-6} M[/tex], V = 325 mL = 0.325 L

[tex]1.8\times 10^{-6} M=\frac{n}{0.325 L}[/tex]

n = [tex]5.85\times 10^{-7} moles [/tex] of iron sulfate.