Question

Calculate the millimoles of solute in 2.25 L of a 0.00645 M NaCN solution. millimoles: mmol NaCN Calculate the millimoles of solute in 275 mL of a 8.89 ppm CaCO 3 solution. millimoles: mmol CaCO 3

Answer 1

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 $CaCO_3$ solution is 0.0244 millimol.

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in L

1) For 0.00645 M NaCN

$0.00645 M=\frac{n}{2.25L}$

$n=0.0145mol=14.5millimoles$    $1mol=10^{3}millimol$

2) For 8.89 ppm

$ppm=\frac{\text {Mass in mg}}{\text {Volume in L}}$

$8.89=\frac{\text {Mass in mg}}{0.275L}$    

${\text {Mass in mg}}=2.44$    

Moles of $CaCO_3=\frac{\text {given Mass in g}}{\text {Molar Mass}}=\frac{2.44\times 10^{-3}}{100g/mol}=2.44\times 10^{-5}mol=0.0244millimol$   $1mol=10^{3}millimol$