Calculate the moles of acetic acid used in each trial and record in Data Table 3.Volume acetic acid in Liters = (Mass/Density)/1000Moles = Volume ∗ Concentration Calculate the moles of magnesium hydroxide (Mg(OH)2) used in each trial and record in Data Table 3.Moles Mg(OH2) = Moles acetic acid
Calculate the neutralization capacity of each trial and record in Data Table. Neutralization Capacity = Moles Mg(OH)2 / Mass of Milk of Magnesium
Data Table Trial 1 Trial 2
Mass of milk of magnesia 2.5g 2.5g
Density of milk of magnesia 1.14 g/mL 1.14 g/mL
Volume of acetic acid, initial 10mL 10mL
Volume of acetic acid, final 2.2mL 1.8mL
Volume of acetic acid, total 7.8mL 8.2mL
Concentration of acetic acid 0.88 M 0.88 M
Moles of acetic acid
Moles of Mg(OH)2
Moles Mg(OH)2 / g milk of magnesia

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.