Question

Part #1:

Which of the following will separate into multiple ions in solution? HCl, NaCl, C6H12O6 (glucose)

A HCl Only

B C6H12O6 Only

C All three

D NaCl Only

E HCl and C6H12O6

F NaCl and C6H12O6

G NaCl and HCl

Part #2:

Determine the ideal van’t Hoff Factor for the following: Al(NO3)3

Part #3:

Determine the molality of total solute particles for the following: 0.227 m Al(NO3)3

Part #4:

What is the freezing point of the solution in part 3 in oC? Note: kf of water is 1.86oC/m and pure water freezes at 0.000 oC (note: as this is freezing point DEPRESSION, what will the final sign of the freezing point be?).

Answer 1

Part #1:

Which of the following will separate into multiple ions in solution? HCl, NaCl, C6H12O6 (glucose)

Solution :-

Answer is option G NaCl and HCl

Because NaCl is salt which is soluble in water which dissociates to form Na+ and Cl- ions

HCl is strong acid which dissociates 100 % to form H+ and Cl-

Part #2:

Determine the ideal van’t Hoff Factor for the following: Al(NO3)3

Solution :-

Dissociation equation for the Al(NO3)3 is as follows

Al(NO3)3(aq) --------- > Al^3+(aq) + 3NO3^-(aq)

On dissociation it gives total 1+3 = 4 ions so the Vant Hoff factor is (i) = 4

Part #3:

Determine the molality of total solute particles for the following: 0.227 m Al(NO3)3

Solution :-

Since Al(NO3)3 gives total 4 ions

Therefore total particle molaity = 0.227 m * 4 = 0.908 m

Part #4:

What is the freezing point of the solution in part 3 in oC? Note: kf of water is 1.86oC/m and pure water freezes at 0.000 oC (note: as this is freezing point DEPRESSION, what will the final sign of the freezing point be?).

Solution :-

Delta Tf= Kf* m

             = 1.86 c/m * 0.908 m

            = 1.69 C

Freezing point of solution = freezing point of water – delta Tf

                                                 = 0 C – 1.69 C

                                                = -1.69 C

So the freezing point of the solution is -1.69 ^oC