Question

will the freezing point of a solution be higher or lower compared to the freezing point of the pure solvent?
question 33
which of the following solutions will have the lowest freezing point?
7.1 m c₂h₃oh solution
7.1 m albr₃ solution
7.1 m nacl solution
7.1 m ca₃(po₄)₂ solution
question 34
which of the following solutions will have the lowest freezing point?
2.6 m c₆h₁₂o₆ solution
1 m mgcl₂ solution
none of these
1 m k₃p
2.6 m nacl solution

Explanation:

First Question (Freezing Point Comparison)

Step1: Recall Colligative Properties

Freezing point depression is a colligative property, meaning it depends on the number of solute particles. For a solution, the freezing point is lower than the pure solvent (freezing point depression: $\Delta T_f = iK_fm$).

Step2: Determine Direction of Change

Since solute particles interfere with solvent crystallization, the solution's freezing point is lower than the pure solvent's.

Step1: Identify Van't Hoff Factor ($i$)

• $C_2H_5OH$ (ethanol): Non - electrolyte, $i = 1$ (does not dissociate).
• $AlBr_3$: Dissociates as $AlBr_3

ightarrow Al^{3 +}+3Br^-$, so $i = 4$.

• $NaCl$: Dissociates as $NaCl

ightarrow Na^++Cl^-$, so $i = 2$.

• $Ca_3(PO_4)_2$: Dissociates as $Ca_3(PO_4)_2

ightarrow 3Ca^{2 +}+2PO_4^{3 -}$, so $i = 5$.

Step2: Relate $i$ to Freezing Point Depression

Freezing point depression $\Delta T_f=iK_fm$. All solutions have the same $m = 7.1\ m$ and $K_f$ (same solvent, assumed). Higher $i$ means higher $\Delta T_f$, so lower freezing point. $Ca_3(PO_4)_2$ has the highest $i = 5$.

Step1: Calculate Effective Particle Concentration ($i\times m$)

• $C_6H_{12}O_6$: Non - electrolyte, $i = 1$, $i\times m=1\times2.6 = 2.6\ m$.
• $MgCl_2$: Dissociates as $MgCl_2

ightarrow Mg^{2 +}+2Cl^-$, $i = 3$, $i\times m = 3\times1=3\ m$.

• $K_3P$: Dissociates as $K_3P

ightarrow 3K^++P^{3 -}$, $i = 4$, $i\times m=4\times1 = 4\ m$.

• $NaCl$: Dissociates as $NaCl

ightarrow Na^++Cl^-$, $i = 2$, $i\times m=2\times2.6 = 5.2\ m$.

Step2: Compare $i\times m$ Values

Higher $i\times m$ means greater freezing point depression (lower freezing point). $NaCl$ solution has $i\times m = 5.2\ m$, which is the highest among the options.

Answer:

lower

Question 33 (Lowest Freezing Point)