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3 years ago

How are the boiling point and freezing point of a solvent affected when a solute is added?

2 answers:

7 0

When a solute is added to a solvent, some properties are affected and these set of properties are called colligative properties. The properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering.

aliya0001 [1]3 years ago

4 0

Answer:

The boiling point increases and freezing point decreases on addition of non volatile solute in a solvent.

Explanation:

When we add a non volatile solute in a volatile solvent, the vapor pressure of the resulting solution is lower than the vapor pressure of pure solvent. Due to decrease in vapor pressure:

a) The temperature required to boil the solution increases (boiling point is the temperature at which the vapor pressure of a liquids is equal to atmospheric pressure). This is known as elevation in boiling point. It is a colligative property which depends upon number of solute particles.

b) The temperature required to freeze a liquid decreases and this is known as depression in freezing point, which is also a colligative property which depends upon number of solute particles.

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3 years ago

What is the pOH of 0.50 molar H3BO3?

Crazy boy [7]

Answer:

A. 10.25

Explanation:

Pkb =4.77

So pka = 14 - pka = 9.23

$Ka =10^{-pka}$

$H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)$

Initial 0.50M 0 0

Change -x +x +x

Equilibrium 0.50M-x +x +x

$Ka =\frac {((x)(x))}{(0.50M-x)}$

$5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}$

(-x is neglected) so we get

$5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}$

$x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}$

$pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76$

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

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