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vesna_86 [32]
3 years ago
5

Chemistry! help! Please!

Chemistry
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

2. a. All three solutes are nonelectrolytes.

3. a. the solution of solute X

Explanation:

The freezing point depression (ΔTf) and <em>boiling point elevation</em> (ΔTb) are <em>colligative properties</em>: they depend on the <em>numbers of particles</em>.

The formula for ΔTf is

ΔTf = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

Sucrose (s) ⟶ sucrose (aq)

1 mole sucrose ⟶ 1 mol particles         i = 1

For NaCl

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol           1 mol     +   1 mol                 i = 2

For Ca(NO₃)₂

Ca(NO₃)₂(s) ⟶ Ca²⁺(aq) + 2NO₃⁻(aq)

1 mol                     1 mol       +  2 mol     i = 3

===============

2. <em>Freezing points </em>

For a nonelectrolyte, i = 1.

Kf = 1.86 °C·kg⁻¹mol⁻¹

b = 1 mol/kg          Calculate ΔTf

ΔTf = 1 × 1.86 × 1

ΔTf = 1.86 °C

Tf = Tf⁰ - ΔTf =0.00 °C – 1.86 °C = -1.86 °C

All the other solutions have lower freezing points, so the solutes must be <em>electrolytes</em>.

===============

3. <em>Boiling points</em>

The formula for ΔTb is

ΔTb = iKb·m

The solution with the <em>highest boiling point</em> will have the <em>highest value of i. </em>

In other words, the solution with the highest boiling point will be the one with the <em>lowest freezing point</em>.

That’s the solution of solute X.

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bearhunter [10]

Answer:

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

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Ka = 1.36 x 10^-3

Explanation:

The reaction

CICH2COOH  ⇄ H+ (aq) + CICH2COO- (aq)

The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.

let us calculate Ka

First , find change in concentration

since , 11% ionized

change in concentration = 0.10 X 11% = 0.011 M

Initial Concentration of CICH2COOH = 0.10 M

change in concentration of CICH2COOH = - 0.011 M

Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

Initial Concentration of CICH2COO- = 0 M

change in concentration of CICH2COO- = + 0.011 M

Equilibrium Concentration of CICH2COO- = 0.011 M

Initial Concentration of H+ = 0 M

change in concentration of H+ = + 0.011 M

Equilibrium Concentration of H+ = 0.011 M

Therefore,

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = [H+][CICH2COO-] /[CICH2COOH]

Ka = (0.011 * 0.011) / (0.089)

Ka = 1.36 x 10^-3

6 0
3 years ago
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