To answer your questions, let's break down the given information and go step by step.
1. To calculate the number of chlorine atoms present in each molecule of A liberated by hydrolysis, we need to use the given data.
We know that 0.805g of compound A was used in the hydrolysis reaction, and 0.718g of silver chloride was formed as a result.
To find the number of chlorine atoms, we will use the molar mass of silver chloride (AgCl). The molar mass of Ag is 107.87 g/mol, and the molar mass of Cl is 35.45 g/mol.
Since the molar mass of AgCl is the sum of the molar masses of Ag and Cl, we can calculate it as follows:
Molar mass of AgCl = Molar mass of Ag + Molar mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Next, we can calculate the number of moles of AgCl using the formula:
Number of moles = Mass / Molar mass
Number of moles of AgCl = 0.718g / 143.32 g/mol = 0.0050 mol
Since AgCl contains one chloride ion (Cl-) per molecule, the number of moles of Cl- will also be 0.0050 mol.
Now, we can calculate the number of chlorine atoms using Avogadro's number (6.022 x 10^23 atoms/mol):
Number of chlorine atoms = Number of moles of Cl- x Avogadro's number
= 0.0050 mol x 6.022 x 10^23 atoms/mol
= 3.011 x 10^21 chlorine atoms
Therefore, there are approximately 3.011 x 10^21 chlorine atoms in each molecule of compound A.
2. The result obtained in question 1 can be explained by understanding the hydrolysis reaction and the formation of silver chloride.
In the reflux reaction between compound A (C7H6Cl2) and excess potassium hydroxide (KOH), the hydrolysis of A takes place. The hydrolysis reaction involves the breaking of chemical bonds using water (H2O).
C7H6Cl2 + 2KOH + H2O → C6H6 + 2KCl + 2H2O
In this reaction, two chloride ions (Cl-) are liberated as potassium chloride (KCl) along with water (H2O) and the formation of compound C6H6 (benzene).
When the resulting solution is acidified with dilute HNO3 acid and excess silver nitrate (AgNO3) is added, a precipitation reaction occurs as silver chloride (AgCl) is formed:
Ag+ (aq) + Cl- (aq) → AgCl (s)
The formation of silver chloride can be used to detect the presence of chloride ions. The stoichiometry of this reaction indicates that one chloride ion (Cl-) reacts with one silver ion (Ag+) to form one molecule of silver chloride (AgCl), which is a white precipitate.
By measuring the mass of the silver chloride formed, we can determine the amount of chloride ions present and, consequently, the number of chlorine atoms originally present in compound A.
Therefore, the result obtained in question 1 suggests that compound A contains two chlorine atoms per molecule, which are liberated as chloride ions during hydrolysis.