Answer: The millimoles of solute in 2.25 L of a 0.00645 M NaCN solution is 14.5 millimoles. The millimoles of solute in 275 mL of a 8.89 ppm [tex]CaCO_3[/tex] solution is 0.0244 millimol.
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in L
1) For 0.00645 M NaCN
[tex]0.00645 M=\frac{n}{2.25L}[/tex]
[tex]n=0.0145mol=14.5millimoles[/tex] [tex]1mol=10^{3}millimol[/tex]
2) For 8.89 ppm
[tex]ppm=\frac{\text {Mass in mg}}{\text {Volume in L}}[/tex]
[tex]8.89=\frac{\text {Mass in mg}}{0.275L}[/tex]
[tex]{\text {Mass in mg}}=2.44[/tex]
Moles of [tex]CaCO_3=\frac{\text {given Mass in g}}{\text {Molar Mass}}=\frac{2.44\times 10^{-3}}{100g/mol}=2.44\times 10^{-5}mol=0.0244millimol[/tex] [tex]1mol=10^{3}millimol[/tex]
