Complete Question
In experiment 1, how many moles of benzoic acid are present? How many moles of sodium bicarbonate are contained in 1 ml of 10% aqueous solution? (A 10% solution has 1 gram of solute in 9 mL of solvent.) Is the amount of sodium bicarbonate sufficient to react with benzoic acid. The amount of Benzoic Acid in ex. 1 is 0.060 grams
Answer:
No of moles of benzoic acid present [tex]n =0.00049 \ moles \approx 0.001 \ moles[/tex]
Number of moles of sodium bicarbonate present [tex]n_s = 0.00119 \approx 0.001[/tex]
Yes the is sufficient sodium bicarbonate to react with Benzoic acid
Explanation:
From the question we are told that
The mass of Benzoic acid is [tex]m_b = 0.060\ g[/tex]
The number of moles of Benzoic acid is
[tex]n = \frac{m_b}{Z_b}[/tex]
where [tex]Z_b[/tex] is the molar mass of Benzoic acid which is a constant with values
[tex]Z_b = 122.12[/tex]
So
[tex]n = \frac{0.06}{122.12}[/tex]
[tex]n =0.00049 \ moles \approx 0.001 \ moles[/tex]
the density of the solution of sodium bicarbonate is
[tex]\rho = \frac{1}{9 }[/tex]
[tex]\rho =0.1 g/mL[/tex]
The mass of sodium bicarbonate is
[tex]m_s = 1 \ mL * 0.1 \ \frac{g}{mL}[/tex]
[tex]m_s = 0.1[/tex]
The molar mass of sodium bicarbonate is a constant with value
[tex]Z_s = 84 \ g/mol[/tex]
The number of moles of sodium bicarbonate is
[tex]n_s = \frac{m_s}{Z_s}[/tex]
[tex]n_s = \frac{0.1}{84}[/tex]
[tex]n_s = 0.00119 \approx 0.001[/tex]
So the number of moles of Benzoic acid and sodium bicarbonate that react are approximately in a 1:1 ratio so we can conclude that there is sufficient sodium bicarbonate to react Benzoic acid
