Answer:
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.
Explanation:
Molarity : It is defined as the number of moles of solute present in one liter of solution. Mathematically written as:
[tex]Molarity=\frac{\text{Moles of solute}}{\text{volume of solution in L}}[/tex]
Moles of citric acid = n
Volume of the citric acid solution = 125 mL =125 × 0.001 L= 0.125 L
(1 mL = 0.001L)
Molarity of the citric acid solution = 0.400 M
[tex]0.400 M=\frac{n}{0.12 5L}[/tex]
n = 0.400 M × 0.125 L = 0.05 moles
Mass of 0.05 moles of citric acid :
[tex]0.05 mol\times 192.12 g/mol=9.606 g[/tex]
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.