Chemistry Concentration Terms
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Class 11 Chapter 1 - Basic Concepts of Chemistry

Concentration Terms

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Reactions in Solutions

`=>` A solution is a homogeneous mixture of two or more substances , the composition of which may vary within limits . " A solution is a special kind of mixture in which substances are intermixed so intimately that they can not be observed as separate components ".

`=>` The substance which is to be dissolved is called solute and the medium in which the solute is dissolved to get a homogeneous mixture is called the solvent.

`=>` In a homogeneous mixture of two substances , the substance which is less in quantity is called solute

In labs, most of the reactions are carried out in solutions. So, amount of substance present in the form of a solution is expressed in the following ways :

Concentration is essentially quantity of solute divided by either the quantity of solvent or solution.

Different Concentration terms differ in the units of numerator ( i.e amount of solute) and denominator ( i.e amount of solvent/solution)

In Denominator we take quantity of solvent, only in case of molality, for other terms we take quantity of solution ( i.e solute + solvent)


(i) Mass percent or weight percent (`w//w %`)

Mass percent `=text( Mass of solute)/text(mass of solution)xx100`

(ii) Mole fraction

Mole fraction of `A` `= text(No. of moles of A)/text(No. of moles of solution)`

(iii) Molarity

Molarity (`M`) `= text(No. of moles of solute)/text(volume of solutions in liters)`

(iv) Molality


Molality(`m`) `= text(No. of moles of solute)/text(Mass of solvent in kg)`

Mass Percent

Mass percent `=text( Mass of solute)/text(mass of solution)xx100`
Q 2683391247

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.

Solution:

Mass per cent of A = `text(Mass of A)/text(Mass of solution ) xx 100`

` = (2 g)/(2g text(of) A +18 g text(of water )) xx 100`

` = (2 g)/(20 g) xx 100`

` = 10%`

Mole Fraction

If a substance `A` is dissolved in substance `B` and their no. of moles are `n_A` and `n_B` respectively, then

Mole fraction of `A` `= text(No. of moles of A)/text(No. of moles of solution)`

Mole fraction of `B` `= text(No. of moles of A)/text(No. of moles of solution)`

Molarity

Definition : It is the number of moles of the solution in `1` L of the solution. It is denoted by `M`.

Molarity (`M`) `= text(No. of moles of solute)/text(volume of solutions in liters)`
Q 2613491340

Calculate the molarity of `NaOH` in the solution prepared by dissolving its `4`g in enough water to form `250` mL of the solution.

Solution:

Since molarity (M)

`= text(No. of moles of solute)/text(Volume of solution in litres)`

`= text(Mass of NaOH/Molar mass of NaOH)/(0.250 L)`

` = (4 g//40 g)/(0.250 L) = (0.1 mol )/(0.250 L)`

` = 0.4 mol L^(-1)`

` = 0.4M`

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent
Q 1913467340

What is the concentration of sugar `(C_12 H_22 O_11)`
in `mol L^(-1)` if its `20 g` are dissolved in enough
water to make a final volume up to `2L`?
Class 11 Exercise 1 Q.No. 11
Solution:

Molar mass of sugar `(C_12 H_22 O_11)`

`= 12xx12+22 xx 1+ 11 xx 16=342 g mol^(-1)`

No. of moles in `20 g` of sugar

`=20/(342 g mol^(-1))=0.0585 mol e`

Molar concentration

`=(text( Moles of solute))/(text(Volume of solution))=0.0585/2=0.0293 M`
Q 1933067842

If the density of methanol is `0.793 kg L^( -1)`, what
is its volume needed for making `2.5 L` of its `0.25`
M solution?
Class 11 Exercise 1 Q.No. 12
Solution:

Molar mass of methanol `(CH_3OH) = 32 g mol^(-1)`

`= 0.032 kg mol^(-1)`

As Molarity `=(text(mass of solute))/(text(molar mass of solute x volume of solution))`

We can write molarity `=(text(density of solution))/(text(molar mass of solute))`

Thus the molarity of the given solutivn will be

`(0.793 kg L^(-1))/(0.032 kg mol^(-1))=24.78 mol L^(-1)`

`24.78 xx V_1 =0.25 xx2.5L`

or `V_1 = 0.02522L=25.22mL`

Molality

Definition : It is the number of moles of the solution in 1 kg of the solvent. It is denoted by `m`.

Molality(`m`) `= text(No. of moles of solute)/text(Mass of solvent in kg)`
Q 1913591440

How are `0.50` mol ` Na_2CO_3` and `0.50 M` `Na_2CO_3`
different?
Class 11 Exercise 1 Q.No. 25
Solution:

Molar mass of `Na_2CO_3 =2 xx 23 + 12 + 3xx16`

`= 106 g mol^(-1)`

`0.50 mol Na_2CO_3` means `= 0.50 xx 10^6 g = 53 g`

`0.50 M Na_2CO_3` means `0.50` mol i.e, `53 g Na_2CO_3`

are present in 1 litre of tile solution
Q 1953280144

A sample of drinking water was found to be
severely contaminated with chloroform, `CHCl_3`,
supposed to be carcinogenic in nature. The level
of contamination was `15` ppm (by mass).

(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in
the water sample.
Class 11 Exercise 1 Q.No. 17
Solution:

(i) 15 ppm means 15 parts in million ( `10^6`) parts,.

% by mass = `15/10^6 xx 100`

`= 15 xx 10^(-4)= 1.5 xx 10^(-3) %`

(ii) Molar mas.; of chloroform

`(CHCl_3)= 2+ 1+3xx 35.5= 119.5 g mol^(-1)`

100 g of the sample contain chloroform

`= 1.5xx 10^(-3) g`

`:. 1000 g (1 kg)` of the sample will contain

Chloroform `=1.5 xx 10^(-2) g`

Now molality

`= (1.5 xx 10 ^(-2))/(119.5)=1.255 xx 10^(-4) m`

Molality `= 1.255 xx 10^(- 4) m`.
 
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