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yuradex [85]
2 years ago
15

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

50) (B) NiSO4 • 6 H2O (M = 263) (C) MgSO4 • 7 H2O (M = 246) (D) Na2SO4 • 10 H2O (M = 286) The correct answer is D, but I really don't know WHY, so I ask... Thank you very much for your help and explanation ! :)
Chemistry
1 answer:
Andrews [41]2 years ago
4 0

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

  • The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

  • Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
  • van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

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The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0
2 years ago
If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the
Kobotan [32]

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

8 0
2 years ago
A sample of CO2 is collected over water at 23oC. If the total pressure of the sample is 734 torr, and the vapor pressure of wate
noname [10]

Answer:

The partial pressure of CO2 is 712,8 in torr

Explanation:

Molar fraction = Pressure in a compound / Total Pressure

Molar fraction H20 = 21,2 / 734 = 0,0288

Sum of molar fraction in a sample = 1

1 - 0,0288 = 0,9712 (molar fraction of CO2)

Molar fraction CO2 = Pressure CO2 / Total pressure

0,9712 . 734 = Pressure CO2

712,8 =Pressure CO2

7 0
2 years ago
Read 2 more answers
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
2 years ago
Pure water is 40 percent acid and 60 percent alkaline <br><br> True or false
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The answer would be true
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