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ratelena [41]
2 years ago
8

Which aqueous solution has the lowest boiling point?

Chemistry
1 answer:
soldi70 [24.7K]2 years ago
8 0

Answer : The correct option is, (a) 1.25 M C_6H_{12}O_6

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

where,

\DeltaT_b = change in boiling point

k_b = boiling point constant

m = molality

i = Van't Hoff factor

As per question, we conclude that the molality of the given solution are same. So, the boiling point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) C_6H_{12}O_6 is a non-electrolyte solute that means they retain their molecularity, an not undergo association or dissociation.

So, Van't Hoff factor = 1

(b) The dissociation of KNO_3 will be,

KNO_3\rightarrow K^++NO_3^-

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(C) The dissociation of KNO_3 will be,

AlCl_3\rightarrow Al^[3+}+3Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The boiling point depends only on the Van't Hoff factor.That means lower the Van't Hoff factor, lower will be the boiling point and vice-versa.

Hence, the correct option is, (a) 1.25 M C_6H_{12}O_6

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Read 2 more answers
Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

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