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Len [333]
3 years ago
8

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. 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
Chemistry
2 answers:
iragen [17]3 years ago
8 0

<u>(NH₄)₃PO₄</u> has  the lowest freezing point.

<h3>Further explanation </h3>

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_fis 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.

<h3>Learn more </h3>

colligative properties

brainly.com/question/8567736

Raoult's law

brainly.com/question/10165688

The vapor pressure of benzene

brainly.com/question/11102916

The freezing point of a solution

brainly.com/question/8564755

brainly.com/question/4593922

brainly.com/question/1196173

atroni [7]3 years ago
4 0

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.

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Answer:

3) About 0.35 grams of hydrogen gas.

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Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To balance it, we can simply add another sodium atom on the left. Hence:

\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}

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\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}

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\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Given the initial value and the above ratios, this yields:

\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Cancel like units:

=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}

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Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

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\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}

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\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

Cancel like units:

\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}

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\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}

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\displaystyle = 65.2 \text{ g Al$_2$O$_3$}

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