Part #1: Which of the following will separate into multiple ions in solution? HC
ID: 913908 • Letter: P
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?).
Explanation / Answer
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