$ \frac {\chi_B}{\chi_A} = \frac {n_B}{n_A} = \frac {\frac {W_B}{M_B} }{\frac {W_A}{M_A}}$
Solved Examples on Molality Formula
Question 1
5% (w/w) glucose solution is given .Find the molarity and molality of solution (density = 1.5gm/cc)
Solution
Molar mass of glucose (C
6 H
12 O
6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol
-1
5% (w/w) glucose solution means 5 gm Glucose in 100 gm solution.
Therefore,Solvent is 95 gm
Now Density = mass /volume
So volume = Mass/Density = 100/1.5 = 1000/15 ml
For molarity:-
M = W/M × 1000/(V ml)
M = 5/180 × (1000/1000) X 15=
= 5/12
Molality (m) = W/MM × 1000/(W gm in solvent)
= 5/180 ×1000/95
= 100/(18 ×19)=50/171
Question 2
A solution of glucose in water is labelled as 10% w/w, Find out
(a)molality
(b) mole fraction of each component in the solution
(c) Molarity
Given the density of solution is 1.2 g m/L
Solution
10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose (C
6 H
12 O
6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g/mol
Then, number of moles of glucose = 10 / 180 mol
= 0.056 mol
Therefore, Molality of solution using Molality Formula= 0.056 mol / 0.09kg = 0.62 m
Number of moles of water = 90g / 18g mol
-1 = 5 mol
Mole fraction of glucose (mg) = 0.056 / ( 0.056+5) = 0.011
And, mole fraction of water mw = 1 - mg
= 1 - 0.011 = 0.989
Now Density = mass /volume
So, volume = Mass/Density
= 100g / 1.2g ml
= 83.33 mL
=83.33 x 10
-3 L
Therefore, Molarity of the solution = 0.056 mol /83.33 x 10
-3 L
= 0.67 M
Question 3
Match the following physical quantities with unit
Physical Quantity
|
Unit
|
(A) Molarity
|
(p) mol
|
(B) Molality
|
(q) Kg
|
(C) Mass
|
(r) mol/kg
|
(d) Density
|
(s) mol/L
|
(e) Mole Fraction
|
(t) g/ml
|
(f) Mole
|
(o) Unit less
|
Solution
Physical Quantity
|
Unit
|
Molarity
|
Mol/L
|
B) Molality
|
Mol/Kg
|
Mass
|
Kg
|
Density
|
g/ml
|
Mole Fraction
|
Unit less
|
Mole
|
Mol
|
Question 4
What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?
(i) 0.1 m
(ii) 1 M
(iii) 0.5 m
(iv) 1 m
Solution
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
No of moles of HCl= 18.25/36.5 =.5 mol
Weight of Solvent= .5 kg
Molality = .5/.5 =1 m
