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Kitty [74]
3 years ago
8

Which of the following scenarios depict an example of colligative properties at play: (A) Pure water freezes at a higher tempera

ture than pure methanol. (B) A 0.5 m NaOH solution freezes at a lower temperature than pure water. (C) A 0.5 m NaBr solution has a higher vapor pressure than a 0.5 m BaCl2 solution.
Chemistry
1 answer:
Veronika [31]3 years ago
6 0

Answer:

(B)

Explanation:

The colligative properties of any given solution are properties that are dependent on the concentration of the molecules or ions of the solute in the solution, and not on the type or identity of that solute. Examples of colligative properties:

1. Lowering of vapour pressure  

2. Elevation of boiling point,

3. Depression of freezing point and

4. Osmotic pressure

(A) is not a colligative property because the scenario talks about the freezing point of a pure solvent and not a solution.

(C) is not a colligative property because the solutions being compared are completely different (the solutes are different).  

(B) is a colligative property because it talks about the freezing point of a solution

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230 cm3 of hot tea at 96 °C are poured into a very thin paper cup with 100 g of crushed ice at 0 °C. Calculate the final tempera
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Answer:

Final temperature of the ice tea, x = 42.73°C

Explanation:

230 cm^3 = 230 g of liquid ( taking the density of tea as the same the density of water)

In cooling from 96°C to 0°

C (before freezing) the available energy is

96 × 230 = 22080 cal

The latent heat (the heat required to melt the crushed ice per gram) of fusion for water = 79.7 cal/g

Melting 100 g of crushed ice at 0°C will absorb

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Or where specific heat of water = 4.186J/Kg°C

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We have

230 × (specific heat capacity of water) × (96 - x) =100× (specific heat capacity of water) ×(0-x) + (latent heat) ×100 =

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Answer:

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Thus the number of electrons in given species are,

Na = 11

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F = 9

F⁻ = 10

Na has more number of electrons.

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