Respuesta :
Solution:
molarity= [tex]\frac{\text{moles}}{\text{volume in litres}}[/tex]
(a) Molarity of NaOH, [tex]M_1[/tex]= 0.03020 M
Volume of NaOH added,[tex]V_1[/tex]= 32.60 mL= 0.0326 L
(1L= 1000mL)
Number of moles of NaOH added :
0.03020 M NaOH means 1L of solution contains 0.03020 moles 0.0326 L contains =[tex]0.03020\times 0.0326[/tex] = 0.00098 moles
(b) Equal number of moles of acid will neutralize equal number of moles of base, thus we can apply molarity equation:
Molarity of citric acid, [tex]M_2[/tex]=?
volume of citric acid, [tex]V_2[/tex]= 20 mL= 0.02 L
using formula, [tex](M_1)(V_1)=(M_2)(V_2)[/tex] ,
[tex]M_2= \frac{(M_1)(V_1)}{(V_2)}[/tex]
[tex]M_2= \frac{(0.03020 M)(0.0326 L)}{(0.02 L)}[/tex] = 0.049 M
(c) Grams of citric acid in 20 ml
moles of citric acid = [tex]{M_2}\times V_1[/tex] of solution in Liters = [tex]0.049 \times 0.02[/tex] = 0.00098 moles
Mass of citric acid =[tex]\text{moles}\times \text{Molecular} mass[/tex]
[tex]= 0.00098\times 192.12[/tex] = 0.1882 g
(d) 20 mL solution of ginger ale contains of citric acid
[tex]0.1882 g= 0.1882 \times 1000 mg[/tex] = 188.2 mg (1g = 1000mg)
Since, 20 mL solution of ginger ale contains of citric acid 188.2 mg
so 1 mL of solution contains = [tex]\frac{188.2 mg}{20 mL}[/tex]
= 9.41 mg/ mL
(a)
Molarity = moles of solute / liter of solution
Moles of solute = molarity x liter of solution
Molarity of NaOH = 0.03020 M
Liter of solution = 32.60 mL / 1000 = 0.0326 L
Moles of NaOH = 0.03020 M x 0.0326 L = 9.85 x 10⁻⁴ moles
(b)
The reaction between NaOH and citric acid is as follows:
C₃H₅O(COOH)₃ + 3NaOH → Na₃C₃H₅O(COO)₃ + 3H₂O
As the balanced chemical equation 3 moles of NaOH reacts with 1 mole of citric acid (Na₃C₃H₅O(COO)₃)
Since the moles of NaOH is 9.85 x 10⁻⁴ the moles of citric acid will be (9.85 x 10⁻⁴) / 3 or 3.284 x 10⁻⁴ moles.
Therefore, the moles of citric acid is 3.284 x 10⁻⁴
(c)
Moles = Mass/ Molar mass
Mass = Moles x Molar mass
Moles of citric acid = 3.284 x 10⁻⁴
Molar mass of citric acid = 192.124 g/mol
Mass of citric acid = 3.284 x 10⁻⁴ x 192.124 g/mol =
Therefore, the mass of citric acid is 0.0631 g
(d)
Mass of citric acid = 0.0631 g
1 g = 1000 mg
Therefore, mass of citric acid = 0.0631 x 1000 = 63.1 mg
(e)
Given,
Volume of citric acid = 20.0 mL
20 ml of citric acid has a mass of 63.1 mg
Therefore, miligram per militier = 63.1 mg/20 ml = 3.155 mg/ml
