Question

question 6 of 29
between a 2.0 m aq. nano₃ and 1.5 m aq. alcl₃ solution:
which would have a lower freezing point?
o 2.0 m nano₃
o 1.5 m alcl₃
o the freezing point is the same for both solutions.
which would have a higher boiling point?
o 2.0 m nano₃
o 1.5 m alcl₃
o the boiling point is the same for both solutions.

Explanation:

Step1: Determine van't Hoff factor (i) for each compound

$NaNO_3$ dissociates into $Na^+$ and $NO_3^-$, so $i = 2$. $AlCl_3$ dissociates into $Al^{3 + }$ and $3Cl^-$, so $i=4$.

Step2: Calculate the effective concentration for freezing - point depression and boiling - point elevation

For $NaNO_3$, effective concentration $= 2.0\ M\times i=2.0\ M\times2 = 4.0\ M$. For $AlCl_3$, effective concentration $=1.5\ M\times i = 1.5\ M\times4=6.0\ M$.

Step3: Analyze freezing - point depression

The greater the effective concentration of solute particles, the lower the freezing point. Since $6.0\ M>4.0\ M$, $1.5\ M\ AlCl_3$ has a lower freezing point.

Step4: Analyze boiling - point elevation

The greater the effective concentration of solute particles, the higher the boiling point. Since $6.0\ M > 4.0\ M$, $1.5\ M\ AlCl_3$ has a higher boiling point.

Answer:

1. B. $1.5\ M\ AlCl_3$
2. B. $1.5\ M\ AlCl_3$