1. Calculate the number of grams of solute required for the preparation of 1.5L of 0.32M NaHCO3 (MW=84) V1 X C1 = V2 X C2

40.32 grams of solute required for the preparation of 1.5L of 0.32M NaHCO₃

Explanation:

Molar concentration or molarity is a measure of the concentration of a solute in a solution. Molarity is defined as the number of moles of solute present in one liter of solution.

Molarity is calculated by the expression:

Molarity is expressed in units .

In this case:

Replacing:

Solving:

number of moles of solute= 0.32 M* 1.5 L

number of moles of solute= 0.48 moles

Being the molar weight of NaHCO₃ equal to 84 g / mole, the following rule of three can be applied: if there are 84 grams in 1 mole, how much mass is there in 0.48 moles?

mass= 40.32 grams

40.32 grams of solute required for the preparation of 1.5L of 0.32M NaHCO₃

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.