What is the percent ionization of H₂NNH₂ in a solution with a concentration of a 0.520 M? (Kb = 1.3 × 10⁻⁶)

% α = 0.1577 %

Explanation:

∴ δH2O = (997 Kg/m³)*(m³/1000 L) = 0.997 Kg/L

∴ C H2NNH2 = (0.520 mol/L)*(L/0.997 Kg) = 0.5216 mol/Kg

∴ Kb = [OH-][H2NNH3+] / [H2NNH2] = 1.3 E-6

⇒ Kb = (X)*(X) / (0.5216 - X) = X²/(0.5216 - X) = 1.3 E-6

⇒ X² + 1.3 E-6X - 6.78 E-7 = 0

⇒ X = 8.227 E-4

⇒ % α = (8.227 E-4 / 0.5216 )*100

⇒ % α = 0.1577 %

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.