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Anna35 [415]
3 years ago
6

Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi

ng compound is dissolved in it. The molar mass of the compound is 70.0 g/mol and the boiling point of the pure carbon disulfide is 46.2°C.
Chemistry
1 answer:
uysha [10]3 years ago
5 0

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

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Answer:

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4 0
3 years ago
Hess’s law
Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

<h3>What is Hess' law?</h3>

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

8 0
2 years ago
How many grams of KCI can be dissolved in 63.5. g of water at 80
Masteriza [31]

Answer:

35.8 g

Explanation:

Step 1: Given data

Mass of water: 63.5 g

Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C

Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.

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8 0
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Chemistry check please... Thank you!
I am Lyosha [343]
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n = 0.024
Your answer is correct.

2) Total pressure = Partial pressure of Hydrogen + Partial pressure of water
134.7  = 122.4 + Pw
Pw = 12.3 kPa
Your answer is correct

3) The molar fraction, volume fraction and pressure fraction of gasses are the same thing.
Thus, percentage pressure of Oxygen = 10%
Pressure of Oxygen = 2.04 x 10⁴ x 0.1
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Well done!
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