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Aleks [24]
3 years ago
11

If 33.9g NaCl are mixed into water and the total mass is 578g, what is the CHANGE in freezing if Kb= - 1.82C/M (molal)? Assume N

aCl does not dissociate in solution.
Chemistry
1 answer:
steposvetlana [31]3 years ago
3 0

Answer:

-1.82 °C

Explanation:

Step 1: Given data

  • Mass of NaCl (solute): 33.9 g
  • Mass of water (solvent): 578 g = 0.578 kg
  • Freezing point depression constant for water (Kb): -1.82 °C/m

Step 2: Calculate the molality of the solution

We will use the following expression.

m = mass of solute / molar mass of solute × kg of solvent

m = 33.9 g / 58.44 g/mol × 0.578 kg

m = 1.00 m

Step 3: Calculate the freezing point depression (ΔT)

The freezing point depression is a colligative property that, for a non-dissociated solute, can be calculated using the following expression:

ΔT = Kb × m

ΔT = -1.82 °C/m × 1.00 m

ΔT = -1.82 °C

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A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

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Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

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1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

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1.8 g of octane produces 3101.56 J of heat

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1.8 g of octane = 3101.56 J

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Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

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