/ Preparation Chemistry Atoms and Molecules
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per litre of solution. Molarity is represented by M. The formula for calculating molarity is given by:
Molarity (M) = Moles of Solute (n) / Volume of Solution (V in litres).
For example, if you have a solution that contains 1 mole of sodium chloride (NaCl) in 1 litre of water, the molarity of the solution would be:
Molarity = 1 mole NaCl / 1 litre H2O = 1 M
Molarity is an important topic in NCERT Class 12 Chemistry. Students will learn this topic in Class 12 Chemistry chapter Solutions. In this article, students will learn molarity definition, formula, difference between molality and molarity etc.
- Molarity: Important terms
- Molarity and Molality: Difference
- FAQs on Molarity
Molarity: Important terms
Molarity (M): This is the concentration of the solute in the solution and is expressed in moles per litre (mol/L).
Moles of Solute (n): This is the amount of the substance being dissolved in the solution, typically measured in moles. You can determine this by weighing the solute and using its molar mass (in g/mol) to convert the mass into moles.
Volume of Solution (V): This is the total volume of the solution and is measured in litres (L).
To calculate molarity, you need to know the number of moles of the solute and the volume of the solution. Once you have these values, you can simply plug them into the formula to find the molarity.
Molarity and Molality: Difference
The difference between molarity and molality is given below.
| Molarity | Molality |
|---|---|
| Molarity of solution is the total number of moles of solute present per litre of solution | Molality is the total moles of a solute present per kilogram of solvent |
| Unit of Molarity is mol L-1 | Unit of molality is mol Kg-1. |
| Change in temperature affects molality | Molality has no effect on temperature |
| Molarity, M = number of moles of the solute /Volume of solution in litres. | Molality, m = Numbers of moles of solute/Mass of solvent in Kg |
| Depends on the volume of the solution | Depends on the mass of the solvent |
FAQs on Molarity
Q: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution
A: 2.5 kg of 0.25 molal aqueous solution.
Molar mass of urea (NH2CONH2) = (2 (1 × 14 + 2 × 1) + 1 × 12 + 1 × 16)
= 60 g/mol
1000 g of water contains 0.25 mol = (0.25 × 60) g of urea.
= 15 g of urea.
Means, 1015 g of solution contains 15 g of urea
Therefore,
2500 g of solution contains = 15 X 2500 / 1015
= 36.95 g
Hence, mass of urea required is 37 g (approx).
Q: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1 .
A: Molality, also called molal concentration, is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. Molar mass of KI = 39 + 127 = 166 g/mol.
20% aqueous solution of KI means 200 g of KI is present in 1000 g of solution. Therefore,
Molality = Moles of KI / Mass of Water in kg
= (200/166) / (0.8) = 1.506 m
(b) Molarity is the concentration of a solution expressed as the number of moles of
solute per litre of solution.
Given,
Density of the solution = 1.202 g/mL
Volume of 100 g solution = mass/ density
= 100/1.202
= 83.19 mL
Therefore, molarity = 20/166 mol/ 83.19 × 10-3 L
= 1.45 M
∴ Molarity of KI = 1.45M
(c) Molar mass of KI = 39 + 127 = 166 g/mol.
Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol
Therefore,
Mole fraction of KI = Moles of KI / Moles of KI+ Moles of Water
= 0.12 / 0.12+ 4.44
= 0.0263
∴ Mole fraction of KI = 0.0263
Q: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant
A:
According to Henry's law,"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,
p = KHx
Where, P = partial pressure of the solute above the solution
KH = Henry's constant
x = concentration of the solute in the solution
Given,
Solubility of H2S in water at STP is 0.195 m
We know,
At STP pressure p = 0.987 bar
0.195 mol of H2S is dissolved in 1000g of water
Moles of water = 1000/18
= 55.56 g/mol
∴ the mole fraction of H2S = Moles of H2S / Moles of H2S + Moles of water
= 0.195 / 0.195+55.6
= 0.0035
According to Henry's law, p = KHx
KH = p/x
KH = 0.987 / 0.0035
KH = 282 bar
∴ The Henry’s law constant is 282 bar
Q: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
A: Given, PAo = 450 mm Hg
PBo = 700 mm Hg
ptotal = 600 mm of Hg
By using Rault's law,
ptotal = PA + PB
ptotal = PAoxA + PBoxB
ptotal = PAoxA + PBo( 1 - xA )
ptotal = (PAo- PBo)xA + PBo
600 = (450 - 700) xA + 700
-100 = -250 xA
xA = 0.4
∴ xB = 1 - xA
xB = 1 – 0.4
xB = 0.6
Now,
PA = PAoxA
PA = 450 × 0.4
PA = 180 mm of Hg and
PB = PBox
PB = 700 × 0.6
PB = 420 mm of Hg
Composition in vapour phase is calculated by
Mole fraction of liquid,
A =PA / PA + PB
= 180/180+420
= 0.30
Mole fraction of liquid,
B =PB / PA + PB
= 420 / 180+420
= 0.70
Q: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2 ) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
A: Given,
Vapour pressure of water, PIo = 23.8 mm of Hg
Weight of water, w1 = 850 g
Weight of urea, w2 = 50 g
Molecular weight of water, M1 = 18 g/mol
Molecular weight of urea, M2 = 60 g/mol
n1 = w1/M1 = 850/18 = 47.22 mol
n2 = w2/M2 = 50/60 = 0.83 mol
We have to calculate vapour pressure of water in the solution p1
By using Raoult's therom,
PI = 23.4 mm of Hg Hence,
The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.
Q: If the density of some lake water is 1.25g mL^–1 and contains 92 g of Na+ ions per kg of water, calculate the molarity of Na+ ions in the lake.
A: Mass of ions = 92g
Molar mass of ions = Na+ = 23g(neglect the mass lost due to absence of a electron)
Moles of ions = mass of ions/molar mass
⇒ n = 92/23 moles
⇒ n = 4moles
Molality of solution = moles of solute/mass of solvent(in kg) Molality = 4/1 = 4M
Q: If the solubility product of CuS is 6 × 10^–16, calculate the maximum molarity of CuS in aqueous solution?
A: The Solubility product of CuS (ksp) = 6 × 10-16
CuS → Cu ++ + S2-
Let the s be solubility of CuS in mol/L
Ksp = [ Cu ++ ][S2]
Ksp = solubility product
6 × 10-16 = s × s = s2
⇒ S = 2.45 × 10-8 mol/L
Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10-8 mol/L
Q: Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
A: Volume of the solution = 250mL = 0.25L
Let the no. of moles of solute be n
Molarity = No. of moles of solute/volume of solution
⇒ 0.15 = n/0.25
⇒ n = 0.0375moles
Molar mass of C6H5OH = 6×12 + 5×1 + 16 + 1 = 94g
Moles = mass/molar mass
⇒ 0.0375 = m/94
Mass of benzoic acid required = 3.525g.
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