Ideally, the van 't Hoff factor should be equal to the number of ions that make up a compound. In reality, van 't Hoff factors tend to be lower due to ion-pairing. Select all conditions that would increase the effect of ion- pairing and decrease the observed van 't Hoff factor. A. Lower charge on the ions
B. High concentration
C. Low concentration
D. Higher charge on the ions

High concentration

Higher charge on the ions

Explanation:

Ideally, the Van't Hoff factor is defined as the ratio between the actual concentration of particles produced when the substance is dissolved in a solution and the concentration of a substance as calculated from its mass(Wikipedia).

The Van't Hoff factor is influenced by the concentration of ions in solution as well as the magnitude of charge on the ions in solution.

Highly charged ions tend to remain paired in solution thereby leading to a lower value of Van't Hoff factor. Also, in highly concentrated solutions, the Van't Hoff factor is lower than in dilute solutions due to the pairing of ions

false

because it does not have the potential to be entirely used up

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g