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galben [10]
3 years ago
11

Calculate the boiling temperature of a 0.50 M solution of sucrose. Assume that the concentration is 0.50 m.

Chemistry
1 answer:
Ronch [10]3 years ago
5 0
There is one missing point in the question.

The formula to find an increase in boiling Temperature is :

ΔT = kb x M

ΔT = is the increase in boiling Temperature
Kb = Boiling point constant of the Solvent
M = Molarity

You did not provide the Kb. If you have it, you just have to insert it to the formula to find the ΔT.


And assuming that the other solution is water, you just have to add it up with 100 Celcius


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Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

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Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

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