Here let me first introduce to the Naming system
Here subscript
A -> Stands for Solvent
B -> Solute
S -> Solution
Solvent (A)
Solute (B)
$W_A$ = Mass of solvent
$M_A$ = Molar mass of solvent
$n_A$ = no. of moles of solvent
$n_A = \frac {W_A}{M_A}$
$W_S$ = Mass of soln
$W_S = W_A + W_B$
$W_B$ = Mass of solute
$M_B $= Molar mass of solute
$n_B$ = no. of moles of solute
$n_B = \frac {W_B}{M_B}$
$d_S $= density of solution
$d_S =\frac { W_S}{V_S}$
MOLARITY:_
Molarity is equal to the no. of moles of solute dissolved in 1 l of solution.
Molarity (M) = $\frac {n_B}{V_S}= \frac {n_B \times 1000}{V_S}$
$= \frac {W_B}{M_B} \times \frac {1000}{V_S}$
Unit of Molarity = Moles/ litre Disadvantage
Molarity of a solution changes with a change in temperature with increase in temperature volume of the soln increases ($V \alpha T$) and therefore molarity decreases.
$M \alpha 1/V$
MOLALITY:-
Molality of a soln is equal to the no. of moles of solute dissolved in 1 kg of solvent.
Molality (m) = $\frac {n_B}{(W_A (kg))}$
Unit of Molality = (m) or moles per kg
Molality of a solution does not change with a change in temperature.
MOLE FRACTION:
$\chi _A = \frac {n_A}{n_A+ n_B }$
$\chi _B = \frac {n_B}{n_A+ n_B }$
The mole fraction of a component is equal to the ratio of no. of moles of that component to the total no. of moles present in the soln.
It is a pure ratio hence it is unit less.
Mole fraction doesn’t change with temperature.
xA + xB = 1.
PERCENTAGE BY MASS OF SOLUTE (%W/W):
% by mass = $\frac {W_B}{W_S} \times 100$ Parts per million:
When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as:
$\text{Parts per million} = \frac {\text {Number of parts of the component}}{\text{Total number of parts of all components of the solution}} \times 10^6$
Solved Examples
Example 1
Four gm of NaOH is present in 100 ml of soln. Find molarity Solution
$M_B = 23 + 16 + 1= 40 $ gm
$W_B = 4$ gm
$V_S = 100$ ml
Molarity = $\frac {W_B}{M_B} \times \frac {1000}{V_S}$
$= \frac {4}{40} \times \frac {1000}{100}$
= 1 M
Example 2
3.42 gm of sucrose ($C_{12}H_{22}O_{11}$) is present in 500 gm water. Find the molality. Solution
For $H_2O$
$W_A = 500$ gm
for $C_{12}H_{22}O_{11}$
$M_B = 342 $ gm, $W_B = 3.42$ gm
Molarity = $\frac {W_B}{M_B} \times \frac {1000}{V_S}$
$= \frac {3.42 \times 1000}{342 \times 500} = 0.02$ m
Example 3
18 gm of glucose is dissolved in 90 gm of water find the mole fraction of glucose & water. Solution
For $H_2O$, $M_A = 18$, $W_A = 90$
for $C_6H_{12}O_6$, $M_B = 180$ gm, $W_B = 18$ gm
$n_B = W_B/M_B =18/180=0.1$
$n_A = W_A/(M_A )=9010/152=5$
$\chi _A = 5/(5+0.1)=50/51$
$\chi _B = 0.1/(5+0.1 )=0.1/5.1=1/51$
Example 4
Calculate molality of 2.5 g of ethanoic acid ($CH_3COOH$) in 75 g of benzene. Solution
Molar mass of $CH_3COOH$: 12 x 2 + 1 x 4 + 16 x 2 = 60 g /mol
Moles of $CH_3COOH$ = 2.5 g/60 g mol= 0.0417 mol
Mass of benzene in kg = 75 g/1000 = $75 \times 10^{–3}$ kg
Molality of $CH_3COOH$ = Moles of $CH_3COOH$/ kg of benzene =$\frac {0.0417}{75 \times 10^{–3}}$
= 0.556 mol/kg
