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Given:

Molarity of HCl, = 0.1 M

Mass of Mixture = 1g

To find: Volume of HCl to react completely with mixture

Formula:

Molarity, M_o = number of moles of solute/ Volume of solution in litres

Solution:

Calculation of Amount of each component in mixture:

⇒ Let the amount of Na₂CO₃ be X g

⇒ And Let amount of NaHCO₃ be [1-X] g

⇒ Molecular Weight of Na₂CO₃ = [23 × 2] + [12] + [3 × 16]

= 106 g

⇒ Molecular Weight of NaHCO₃ = [23] + [1] + [12] + [3 × 16]

= 84 g

⇒ Number of moles of NaHCO₃ = 1-x / 84

⇒ Number of moles of Na₂CO₃ = x / 106

Now it is given in the question that the mixture is equimolar, so

⇒ Number moles of Na₂CO₃ = Number of moles of NaHCO₃

1-x/84 = x/106

106 – 106X = 84X

190X = 106

x= 106/190

X = 0.5579

≈ 0.558

Amount of Na₂CO₃ = 0.558 g

⇒ Amount of NaHCO₃ = 1- 0.558 g = 0.442 g

⇒ Number moles of Na₂CO₃ = [0.558/106]

= 0.00526 moles

⇒ Number of moles of NaHCO₃ = 0.00526 moles …. [Mixture is equimolar]

Calculation of volume of HCl required

⇒ In the question it is given that HCl reacts with the mixture.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

NaHCO₃ + HCl → NaCl + H₂O + CO₂

From the above chemical equations it can be found that 2 moles of HCl is required to react with one mole of Na₂CO₃ and 1 moles of HCl is required to react with one mole of NaHCO₃.

So total number of moles of HCl required to completely react with mixture is as follows:

⇒ Total Number of Moles of HCl = [2 × 0.00526] + 0.00526

= 0.01578 moles

From Molarity Formula we have,

Molarity, M_o = number of moles of solute/ volume in litres (V)

⇒ Volume in litres, V = [0.01578/0.1]

= 0.1578 L

⇒ Volume in ml, V = 0.1578 × 1000

= 157.8 ml

Therefore the volume of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both is 157.8 ml.

Given:

10% w/w solution means that if solution is of 100g then 10g of glucose is present in it [10 × 100]/100] and the amount of water present in it is [100-10], M_w = 90g.

Density, d = 1.2g/ml

To find: Molality and Mole Fraction of Each component Formula:

Density, d = Mass (M) /volume (V)

Molality = Number of moles of solute/ Mass of solvent in kg

Mole Fraction of component  = number of moles of given component / total number of moles in solution

Molarity, M_o = Number of moles of solute/ Volume of solution in litres

Solution:

Calculation of Molality:

⇒ Molecular mass of Glucose, [C₆H₁₂O₆] = [6 × 12] + [12 × 1] + [6 × 16]

= 72 + 12 + 96

= 180g

⇒ Number of moles of Glucose, N_g = [10/180]

= 0.0556 moles

⇒ Molality = N_g/M_W

⇒ Molality = 0.0556/ 0.09

= 0.6177 m

≈ 0.62 m

Calculation of Mole Fraction:

⇒ Molecular Mass of Water, [H₂O] = [2 × 1] + [16 × 1]

= 2 + 16

= 18g

⇒ Number of moles of water, N_w = [90/18]

= 5 moles

⇒ Mole Fraction of Glucose,

x_g = N_G / N_W + N_G

= 0.0556 / 5+0.0556

= 0.011

⇒ Mole Fraction of Water, x_w = (1-x_g)

= 1 – 0.011

= 0.989

Calculation of Molarity:

From density we can find out volume.

⇒ Volume, V = [100/1.2]

= 83.33 ml

⇒ Molarity  = 0.0556/ 83.33 X 1000

Molarity, M = 0.667 M

≈ 0.67 M

Therefore the answers obtained in the numerical can be summarized in following table:

Given:

Concentration of Nitric Acid, HNO₃ = 68%

Density of solution, d = 1.504 g/ml

To find: Molarity, M_o

Formula:

Density, d = Mass (M) / volume (V)

Molarity, M_o  = Number of moles of solute/ Volume of solution in litres

Solution:

68% of Nitric acid by mass in aqueous solution means that 68g [68 × 100]/100] of Nitric acid present in 100g of solution.

⇒ Molecular mass of Nitric Acid, HNO₃ = [1 × 1] + [1 × 14] + [16 × 3]

= 63g

⇒ Number of moles of Nitric Acid = [68/63]

= 1.079 moles

⇒ Given Density, d = 1.504 g/ml

⇒ Volume, v = [100/1.504]

= 66.489 ml

⇒ Molarity, M_o = [1.079/66.489] × 1000

= 16.23 M

Therefore the molarity of the sample is 16.24 M.

(1) Mole fraction - The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution.

Mathematically,

Mole Fraction of component = Number of moles of given component / Total number of moles in solution

Mole Fraction is independent of temperature.

(2) Molality - Molality of a solution is defined as the number of moles of solute dissolved per 1000g [1kg] of the It is represented by m.

Molality actually represents the concentration of solution in mol / kg.

Mathematically,

Molality = Number of moles of solute/ Mass of solvent in kg

It is represented by m.

(3) Molarity- The number of moles of solute dissolved per litre of the solution at a particular temperature is called the molarity of the solution at that

Molarity actually represents the concentration of a solution in mol / L.

Mathematically,

Molarity = Number of moles of solute/ Volume of solution in litres

(4) Mass percentage - Mass percentage is defined as the mass of the solute in grams dissolved per 100g of the It is also referred to as weight percentage [w/w].

For example, 10% [by mass] urea solution means that 10 g of urea are present in 100 g of solution, the solvent being only 100-10 = 90 g.

Mathematically, the mass percentage of a solute in a solution is given by:

Mass Percentage of Solute = Mass of solute / Mass of solute + Mass of solvent X 100

Or

Mass Percentage of Solute = Mass of solute / Mass of solution X 100

The fee structure designed for SRM University BCom courses is a composition of various fee elements to be paid by the students to complete the course. According to the fee structure, students are required to pay two major components: tuition fees and hostel fees.

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