I believe the answer is a definition.
Answer:
2.93g
Explanation:first, let us calculate the number of mole of NaCl present in the solution. This is illustrated below:
Molarity = 0.5M
Volume = 100cm^3 = 100/1000 = 0.1L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of NaCl = 0.5 x 0.1 = 0.05mole
Now we can obtain the mass of NaCl as follows:
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mole of NaCl = 0.05mol
Mass of NaCl =?
Mass = number of mole x molar Mass
Mass of NaCl = 0.05 x 58.5
Mass of NaCl = 2.93g
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
To measure weight of a item
