In our everyday life, we come across various solutions. Their use or importance in life depends on their composition. So what is mean by solutions?
Solutions
Solutions are
said to be homogenous mixtures of different components. Their composition and properties
are uniform throughout the solution. The components of a solution include,
1)
Solvent: solvent will be present in
large quantities and it will determine the physical state of a solution.
2)
Solutes: they are the components
present in the solution other than solvent.
There are
different types of solutions present based on the state of solvent. If the
solvent is a gas, then the solution is said to be gaseous. Liqyiud solutions
contain liquid solvent and solid solutions contain solid solvent.
How do we
express the compostion of solutions
We can express
the composition qualitatively as well as quantitatively.
There are
several tools available to express the composition quantitatively and they are
·
Molarity
·
Molality
·
Mass percentage
·
Volume percentage
·
Mass by volume percentage
·
Parts per million
·
Mole fraction
Molarity(M)
It is defined as
number of moles of solute dissolved in one litre of solution
Molarity= Moles
of solute/ volume of solution in litre
For example, 0.3
M solution of NaCl means that 0.3 mol of NaCl has been dissolved in one litre.
Now, solve some
problems.
Download the questions in PDF format, Link is given below
1.
Calculate the mass of sodium acetate (CH3COONa)
required to make 500 ml of 0.375 molar aqueous solutions. The molar mass of sodium
acetate is 82.0245 g mol-1
A)
12.31
B)
22.18
C)
14.30
D)
15.38
Answer: D
Molarity= No. of moles of solute/volume of solution in
litre
Molarity = 0.375 M
No. of Moles=? (Mass of the substance/ Mol. Mass of the
substance)
The molar mass of Sodium acetate = 82.0245 g Mol-1
Volume of solution= 500 ml
Converting to litre= 500/1000=0.5
Substitute the values,
0.375 M= (mass of substance/82.0245)/0.5
0.375X0.5X82.0245=15.38
The mass of sodium nitrate required is 15.38
2.
The concentrated H2SO4 that is paddled
commercially is 95% H2SO4 by weight. If the density of this commercial acid is
1.834g cm-3, the molarity of the solution is?
A)
17.8M
B)
12.62M
C)
19.04M
D)
12.07M
Answer: A
Molarity= No. of moles of solute / volume of solution
95% means, 100g contains 95g H2SO4
Mol. Mass of H2SO4 = (8.08 g/mol
Moles of H2SO4= 95/98.08= 0.968 mol
Density= mass/volume
Volume of solution = mass/density
=
100g/1.834
= 0.05452 L
Molarity= moles
of H2SO4/volume of solution in litres
=
0.968 mol/0.05452 L
=
17.8 M
3.
Calculate the concentration of nitric acid in
moles per litre in a sample which has a density 1.41 g mL-1 and the
mass percentage of nitric acid in it being 69%.
A)
13.04
B)
14.29
C)
15.44
D)
19.42
Answer: C
69% means 100g contains 69g of HNO3
Volume of solution= mass/ density
= 100g/1.41g mL-1
= 70.92
Converting ml to L= 70.92/1000=0.07092
Moles of HNO3 = 69/63= 1.095
Molarity= 1.095/0.07092
= 15.44 M
4.
What is the mass of NaCl required to make 750
mL of 0.25M solution in water.
A)
11.56
B)
10.95
C)
16.71
D)
15.88
Answer: B
Molarity= No. of Moles of solute/volume of solution
Volume in Litre= 750ml=0.75L
No. of moles =? /58.44
Mass required= 0.25x0.75x58.44=10.95
10.95 g
5.
What is the concentration of sodium tartrate
(C4H4NaO6) in mol L-1? It its 50g are dissolved in water to make
final volume up to 3L?
A)
0.08
B)
0.12
C)
0.03
D)
0.91
Answer: A
The molecular mass of Sodium tartrate = 194.051 g mol-1
Moles of Sodium tartrate = 50/194.051
=
0.25 mol
Volume of solution = 3
Molarity= 0.25/3 = 0.08 M
6.
How many grams of CaCl2 are
contained in 250ml of 0.3M CaCl2 solution?
A)
4.37
B)
7.21
C)
7.98
D)
8.3
Answer: D
Mol. Mass of Cacl2= 110.98
Molarity = 0.3
Volume = 0.25 L
Mass required= 0.3x 0.25x 110.98= 8.3g
7.
The density of a solution prepared by
dissolving 120g of urea (mol mass= 60 u) in 1000 g of water is 1.15 g/ml. the molarity
of this solution is?
A)
1.04
B)
3.08
C)
2.05
D)
1.88
Answer: C
Molarity= Number of moles of urea/ volume of solution
Volume of solution= mass of solution/density
= (120+1000)
g/1.15g mL-1
= 974ml=0.974L
Molarity= 2/0.974= 2.05 mol L-1
8.
What is the concentration of sugar (C12H22O11)
in mol L-1? If its 20g are dissolved in enough water to make a final volume up
to 2L?
A)
0.029
B)
0.054
C)
0.042
D)
0.013
Answer: A
The mol. Mass of sugar= 342
Moles of sugar= 20/ 342=0.058
Volume of solution= 2L
Molarity= moles of solute/ volume of solution in litre=
0.058/2
= 0.029 mol/L
9. If the density of methanol is 0.793 Kg/L,
what is its volume needed for making 2.5 L of its 0.25 M solution?
A)
25.9
B)
24.07
C)
25.2
D)
27.1
Answer: C
Molarity= moles of CH3OH/volume in litre
0.25= moles of CH3OH/2.5
Moles of CH3OH=2.5X0.25= 0.625
Mass of CH3OH= 0.625X32=20g
0.793x103g of CH3OH is present in 1000Ml
20g of CH3OH is present in= (1000/0.793x103)
x20
= 25.2 ml
10. How many grams of HCl are required to prepare
4L of 5M HCl in the water?
A)
750g
B)
730g
C)
780g
D)
710g
Answer: B
Molarity= moles of solute/volume in L
No of moles of HCl = mass/mol. Mass
5M= (mass of HCl/36.5)/4
Mass of HCl required= 5x20x36.5
= 730g
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