Chemistry Concentration Terms

### 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 4g 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.