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

Question

asked Aug 12, 2018 197k views

2 Answers

Mdsingh · Answer 1 · 2018-08-19T13:03:53+0000

Answer:

Step-by-step explanation:

Depression in freezing point is given by:

$\Delta T_f=i* K_f* m$

$\Delta T_f$ = Depression in freezing point

i= vant hoff factor

K_f = freezing point constant

m= molality =
$\frac{moles}{\text {weight of solvent in kg}}=(1)/(1)=1m$

1. For
PbS

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

2. For
C_8H_(18)

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

3. For
Na_3PO_4

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

$Na_3PO_4\rightarrow 3Na^(+)PO_4^(3-)$

4. For
NaCl

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

$NaCl\rightarrow Na^(+)+Cl^(-)$

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