#### Lesson / Chemistry / Liquid State

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### Notes on Liquid State

216 / May-30-2021 12:37:16 By- srijan /

1. Solubility: The amount of solute dissolve in 100 gm of solvent to make its saturated solution at a particular temperature is called solubility.
Solubility= (Weight of solute/Weight of solvent) x 100%
2. Solubility Curve: The curves obtained by plotting the solubility of substance in Y-axis and temperature in X-axis are called solubility curve.
There are two types of solubility curves:-
i. Continuous solubility curve: The curves in which there is regular increases or decreases the solubility of the substance with the temperature is called continuous solubility curve.
ii. Discontinuous solubility curve: The curves in which there is irregular increases or decreases the solubility of the substance with the temperature is called discontinuous solubility curve.

## Application of solubility curve or information obtained from a solubility curve:-

1. It explains the effect of temperature on the solubility of the substance.
2. It indicates the solubility of a substance at a particular temperature.
3. It informs about the nature of substance whether it is hydrous or anhydrous.
4. It shows a change in the composition of the solute.
5. It helps in comparing the solubility of different substances. During fractional crystallization we can predict by solubility curve which subtends in binary mixture.

### Viscosity

The property of liquid by virtue of which its one layer resists the flow of another layer is called viscosity.
It is the frictional force between the two layers of liquid, the force is called viscous force. We know that,
F Adv/dx
or, F= Adv/dx , where
F= viscous force
A= area of layers
dv= velocity differences between two layers of liquid.
dx= distance between the layers.
= coefficient of viscosity.
when, A= 1cm2, dv= 1cm/sec and dx= 1cm then,
F=
Thus, the coefficient of viscosity is the force per unit area required to maintain the unit difference of velocity between the two layers of liquid in unit distance apart.

Avogadro's hypothesis and its application

It states that, "Equal volume of all gases contains equal number of molecules under similar condition of temperature and pressure."
Let, V1 volume of gas A contain n1 molecule and V2 volume of gas B contain n2 molecule.
If V1=V2 then from Avogadro's hypothesis,
n1=n2
# Application of Avogadro's hypothesis:
1. Determination of atomic weight of elements.
2. Determination of Molecular formula of gaseous substance from their volumetric composition.
3. Determination of mass-volume relationship of gases.
4. Determination of relationship between the molecular mass and vapour density.
5. Determination of atomicity (no. of atoms) of elemental gases.

Vapour density:

It is the ratio of weight of certain volume of gas to the weight of same volume of hydrogen gas at constant temperature and pressure.

i.e Vapour Density (V.D)= wt. of certain volume of gas/wt. of same volume of H2 gas

Relationship between Molecular Mass and Vapour Density:

We know that,
V.D= wt. of certain volume of gas/wt. of same volume of H2 gas [at constant temperature and pressure]
or, V.D= wt. of 'V' volume of gas/wt. of 'V' volume of H2 gas [at constant temperature and pressure]
Let, 'V' volume gas contain 'n' molecules. Then apply avogadro's hypothesis,
V.D= wt. of n molecules of gas/wt. of n molecules of H2 gas [at constant temperature and pressure]
or, V.D= wt. of 1 molecules of gas/2 x wt. of 1 atom of hydrogen [at constant temperature and pressure]
2V.D= Molecular weight
therefore, Molecular weight= 2 x V.D
where, Molecular mass= wt. of 1 molecule of gas/wt. of atom of hydrogen

Some numericals:

Q. Calculate the weight of crystal formed on cooling 80gm of saturated solution from 60 degree celsius to 30 degree celsius. Solubility of salt at 60 degree celsius and 30 degree celsius are 132 and 95 respectively.
Let, A and B be the weight of salt at 60 degree celsius and 30 degree celsius respectively.
For 600C
weight of salt= A
weight of solution= 80 gram
solubility= 132
weight of solvent= (80-A)gram
we have
solubility= (wt.of salt/wt.of solvent) x 100
132= (A/80-A) x 100
or, 10560-132A= 100A
or, 10560= 232A
A= 45.52 (weight of salt)
therefore, weight of solvent= 80-A
=34.48 gram

For 300C
weight of salt= B
weight of solvent= 34.48 gram
solubility= (wt.of salt/wt.of solvent) x 100
95= (B/34.48) x 100
327506= 100B
therefore, B= 32.76
Therefore, the weight of crystal formed= (A-B)
= 12.76 gram.

Q. 1 litre of hydrogen at STP weights 0.09 gram. If 2 litres of a gas at STP weights 2.880 gram, calculate the vapour density and the molecular weight of the gas.
solution:
wt. of 1 litre of H2 at STP= 0.09 gram
wt. of 2 litre of a gas at STP= 2.880 gram
wt. of 1 litre of a gas at STP= 2.880/2 gram = 1.440 gram
vapour density (v.d)= (wt. of certain volume of gas/wt. of same volume of H2 gas) [At constant temperature and pressure]
or, v.d= 1.440/0.09
therefore, v.d= 16
we have,
molecular weight= 2 x v.d
= 2 x 16
= 32