Solutions

Given:

Mass of ethylene glycol (C₂H₆O₂) = 222.6 g

Mass of water = 200 g

Density, d = 1.072 g/ml

To find: Molality and Molarity of solution

Formula:

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

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

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

Calculation of Molality:

⇒ Molecular Mass of ethylene glycol (C₂H₆O₂) = [12 * 2] + [6 * 1] + [16 * 2]

= 24 + 6 + 32

= 62 g

⇒ Number of moles of ethylene glycol (C₂H₆O₂) = [222.6/62]

= 3.59 moles

⇒ Mass of water = 200 g

⇒ Molality = numebr of moles of solute/ Mass of solvent in kg

= 3.59 / 200 X 1000

= 17.95 m

Calculation of Molarity:

⇒ Total mass of solution = [222.6 + 200]

= 422.6 g

From density formula we can find out the volume required.

⇒ Volume of solution, V = [422.6 /1.072]

= 394.216 ml

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

⇒ Molarity, M_o =3.59 / 394.216 X 1000

⇒ Molarity = 9.1067

⇒ Molarity ≈ 9.11 M

Therefore the Molality and Molarity of the solution is as follows:

Molality = 17.95 m

Molarity = 9.11 M

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:

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.