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alexdok [17]
3 years ago
9

The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van

't hoff factor for this solute if the kb for water is 0.512°c/m. 2.7 1.8 2.3 3.0 3.6
Chemistry
1 answer:
Goshia [24]3 years ago
6 0

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: ΔT_{b}=i K_{b}.m, where ΔT_{b}= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

K_{b}= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=\frac{2.5}{0.512 X 1.83}

i=2.66= 2.7 (approx.)


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The substances below are listed by increasing specific heat capacity value. Starting at 30 Celsius, they absorb 100 kJ of therma
Georgia [21]

Answer:

Silver.

Explanation:

To obtain the right answer to question, let us calculate the change in temperature for each substance assuming they all have the same mass as 100g.

This is illustrated below:

1. For Siver:

Mass (M) = 100g

Specific heat capacity (C) = 0.239J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.239)

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2. For Aluminium:

Mass (M) = 100g

Specific heat capacity (C) = 0.921J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.921)

ΔT = 1086°C

3. For Lithium:

Mass (M) = 100g

Specific heat capacity (C) = 3.56J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 3.56 )

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4. For water:

Mass (M) = 100g

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 4.184)

ΔT = 239°C

Summary

Temperature change of each substance is given below

1. Silver => 4184°C

2. Aluminum => 1086°C

3. Lithium => 281°C

4. Water => 239°C

From the calculations made above, Silver has the highest rise in temperature.

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<em>V= 110mL = 110cm³ = 0,11dm³</em>
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C = n/V
n = 1,244×0,11
<u>n = 0,13684 moles</u>

<em>mCa(OH)₂ = 74 g/mol</em>


1 mole Ca(OH)₂ ------------ 74g
0,13684 ---------------------- X
X = 74×0,13684
<u>X = 10,12616g</u>

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