Which substance will most affect the freezing point of water if one mole of the substance is added to one kilogram of water?

pbs

c8h18

na3po4

nacl

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

i= vant hoff factor

= freezing point constant

m= molality =

1. For

, i= 1 as it is a non electrolyte and does not dissociate.

2. For

, i= 1 as it is a non electrolyte and does not dissociate.

3. For

, i= 4 as it is a electrolyte and dissociate to give 4 ions.

4. For

, i= 2 as it is a electrolyte and dissociate to give 2 ions.

Thus as vant hoff factor is highest for and the freezing point will be most affected.

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

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.