Question

assume ideal van't Hoff factors where applicable. Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.075 m KNO2 0.075 m LiCN 0.075 m (NH4)3PO4 0.075 m NaI 0.075 m NaBrO4

Answer 1

(NH₄)₃PO₄ has  the lowest freezing point.

Further explanation

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.

Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles because electrolytes break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes.

The term is used in the Solution properties

• 1. molal

that is, the number of moles of solute in 1 kg of solvent

$\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}$

• 2. Boiling point and freezing point

Solutions from volatile substances have a higher boiling point and lower freezing points than the solvent

ΔTb = Tb solution - Tb solvent

ΔTb = boiling point elevation

$\rm \Delta T_f=T_fsolvent-T_fsolution$

$\large {\boxed {\boxed {\bold {\Delta Tb \: = \: Kb.m}}}$

$\rm \Delta T_f=K_f\times m$

Kb = molal boiling point increase

Kf = molal freezing point constant

m = molal solution

For electrolyte solutions there is a van't Hoff factor = i

i = 1 + (n-1) α

n = number of ions from the electrolyte

α = degree of ionization, strong electrolyte α = 1

so the freezing point formula becomes:

$\rm \Delta T_f=K_f\times m\times i$

All solutions in the problem have molal concentration (0.075 m) and the same solvent -> assuming water (The same $\rm K_f$) so that what affects the value of $\rm \Delta T_f$is the value of i

Assuming the degree of electrolyte ionization α= 1, the magnitude i is determined by the number of ions produced by the electrolyte (n)

KNO₂ ---> K⁺ + NO₂⁻ → 2 ions

LiCN ---> Li⁺+ CN⁻ → 2 ions

(NH₄)₃PO₄---> 3 NH₄ + + PO₄ ³⁻ → 4 ions

NaI ---> Na⁺ + I⁻ → 2 ions

NaBrO₄ ---> Na⁺ + BrO₄⁻ → 2 ions

(NH₄)₃PO₄ has the highest number of ions, so it has the highest $\rm \Delta T_f$ and the lowest freezing point.

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Answer 2

Answer:

$(NH_4)_3PO_4$ 0.075 m solution has the lowest freezing point.

Explanation:

Depression in freezing point is given by:

$\Delta T_f=T-T_f$

$\Delta T_f=K_f\times m$

$\Delta T_f=iK_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$

where,

$\Delta T_f$ =Depression in freezing point

$K_f$ = Freezing point constant of solvent

1 - van't Hoff factor

m = molality

According question, molality of all the solutions are same and are in prepared with same solvent. So, values of molality and $K_f$ will remain the same and will not effect the freezing point of the solution.

The lowering in freezing point will now depend upon van't Hoff factors of the solutions. Higher the value of van'Hoff factor more will be the lowering in freezing point of the solution.

The van't Hoff factor of $KNO_2$ solution = $i_1=2$

The van't Hoff factor of $LiCN$ solution = $i_2=2$

The van't Hoff factor of $(NH_4)_3PO_4$ solution = $i_3=4$

The van't Hoff factor of $NaI$ solution = $i_4=2$

The van't Hoff factor of $NaBrO_3$ solution = $i_5=2$

The solution of ammonium phosphate has the highest values of van't Hoff factor which will result in maximum lowering of the freezing point of the solution.

Hence,$(NH_4)_3PO_4$ 0.075 m solution has the lowest freezing point.