Question

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.

Answer 1

Explanation:

$c=\frac{n}{V}$

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

$18.5 M=\frac{n}{2.00 L}$

n = 37 moles of sulfuric acid

b) 100.0 mL of $3.8\times 10^{-5} M$ of  sodium cyanide

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

$3.8\times 10^{-5} M=\frac{n}{0.1 L}$

n = $3.8\times 10^{-6} moles$ of sodium cyanide

c) 5.50 L of 13.3 M of concentrated formaldehyde.

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

$13.3 M=\frac{n}{5.50 L}$

n = 73.15 moles of formaldehyde.

d)325 mL of $1.8\times 10^{-6} M$ of  iron sulphate

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

$1.8\times 10^{-6} M=\frac{n}{0.325 L}$

n = $5.85\times 10^{-7} moles$ of iron sulfate.