It is defined as the number of moles of solute dissolved per litre of solution. It is expressed by the symbol M.Its units are mol/L.
Here MB is molecular mass and WB is the given mass of solute.It is the most common and widely used method in calculations regarding concentration of solutions in laboratory.However Molarity has a disadvantage.It changes with temperature due to expansion and contraction of liquid solutions with variable temperatures
MOLAR SOLUTION
When one mole of the solute is present in one litre of the solution.Then the solution is called one molar solution.For example when 126 gram of oxalic acid is dissolved in one litre of solution,it is one molar solution as 126 gram is the molar mass of the oxalic acid.
NUMERICALS ON CONC.
1)Calculate the concentration of nitric acid in moles per liter in a sample which has a density of 1.41 g/ml and mass percent of nitric acid in it being 69%.(Ans:-15.44 M)
2)2.46 g of NaOH (Molar mass = 40g/mol) has been disolved in water to make the solution 100 ml.Calculate Molarity.(Ans:0.615 M or 0.615 mol/L)
3)A solution of oxalic acid(Molar mass=126 g/mol) is prepared by dissolving 0.63g of acid in 250 ml of solution .Calculate Molarity.(Ans:0.02 M)
4)A solution contains 38% HCl by volume.Density of solution is 1.19g/cm3.Calculate Molarity.(Ans:12.39 M)
5)A solution contains 38% HCl by mass.Density of solution is 1.19g/cm3.Calculate Molarity.(Ans: 12.39 M)
6)The 10% solution of KCl by mass has density 1.06g/cm3.Calculate Molarity .Molar mass of KCl = 74.5 g/mol.(Ans:1.42 M)
7)Caculate Molarity of 15% sulphuric acid (w/w).Whose density is 1.02 g/cm3.Molar mass of sulphuric acid = 98g/mol.
8)Calcualte Molarity of solution containing 4 gram of NaOH(Molar mass =40 gram/mol) dissolved in 200 ml of solution.(Ans:0.5 M)
9)A solution has been prepared by dissolving 18.25 g of NaOH in water.Volume of solution is 340 ml.Find Molarity.(Ans:1.342 M)
10)0.38 gram of NaNO3has been dissolved in 250 ml of solution.Find out Molarity.(Ans:1.78*10-2M).
11)0.016 moles of Glucose (molar mass = 180 g/mol)are present in 250 ml of solution. Find out Molarity.(Ans:0.063 M)
12)Calculate amount of benzoic acid required for preparing 250 ml of 0.15 M sol.(Ans: 4.575 gram)
13)How can you prepare one molar solution of sulphuric acid in water?
Ans:As one molar sulphuric acid weighs 98 gram hence Just dissolve 98 gram of sulhuric acid in some water and make the final volume of solution 1000 ml
14)How can you prepare one molar solution of oxalic acid?
Ans:Just dissolve 126 gram of oxalic acid in some water and make the final volume of solution 1000 ml
15)How can you prepare of M/20 solution of Oxalic acid?
Ans:Just take the twentieth part of molar mass of oxalic acid that is 126/20 = 6.3 gram.
Now dissolve 6.3 gram of of oxalic acid in some water and make the final volume of the solution equal to 1000 ml.
16)How would you prepare 250 ml of M/20 solution of oxalic acid?
Ans:Again divide the molar mass of oxalic acid that is 126 gram by 20. 126/20 20 6.3 gram.Now this amount is for one litre solution and we are asked to make the final volume of the solution 250 ml.Hence we will divide 6.3 gram with 4, 6.3/4 = 1.575 gram.Now we will dissolve 1.575 gram of oxalic acid in water and will make the final volume of lthe solution exactly 250 ml.
17)A solution of glucose in water is 10% by mass.Density of solution is 1.20g/cm3.Find Molarity.(Ans:0.667 M)
222.6 gram of ethylene was dissolved in 422.6 gram of solution.Density of sol is 1.072 g/cm3.Calculate Molarity.(Ans:9.11 M)
18)Calculate volume of sulphuric acid required to prepare one litre of 0.2 M sulphuric acid .You are given 80 gram of sulphuric acid present in 55.5 ml of solution.
First calculate molarity of sulphuric acid which comes out to be = 14.7 M. Then apply moalrity equation to find out volume
(Ans:Moalrity of sulphuric acid:14.7 M and upon applying Molarity equation we get volume required = 13.60 ml)
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