Solutions

NCERT Solution part-2

Question (4) Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?

Solution

Given, concentration of nitric acid by mass in aqueous solution in % = 68%

And Density of the given solution of nitric acid = 1.504 gmL^–1

Thus, Molarity = ?

Here the meaning of 68% of nitric acid solution by mass is 68 g of nitric acid is dissolved in 100 g of solution by mass.

And, Molar mass of nitric acid (HNO₃)

= 1 + 14 + 3xx16=15+48=63 g mol^-1

Therefore, moles of 68 g of nitric acid (HNO₃)

`=68/63` =1.079 moles

Now, we know that volume of a solution = mass/density

Thus, volume of given solution `=100/1.504\ mL`

= 66.5 mL

`=66.5/1000` litre

Thus, volume of given solution in litre = 0.0665 litre

Now, we know that Molarity of solution

= moles of solute/Volume of solution in litre

Thus, molarity of given solution of nitric acid

`=1.079/0.0665` = 16.225 M Or 16.23 M

Thus, molarity of given solution of nitric acid = 16.23 M Answer

Question (5) A solution of glucose in water is labeled as 10% w/w. What would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^-1, then what shall be the molarity of the solution?

Solution

Given, Concentration of solution of glucose in water = 10% w/w

Density of solution = 1.2 g mL^–1

Thus, Molality, mole fraction and molarity = ?

10% of solution of glucose in water means solution contains 10g of glucose and 90 g of water.

Molar mass of glucose (C₆H₁₂O₆)

=6 × × 12 + 12 × × 1 + 6 × ÷ 16

=72+12+96 = 180 g mol^-1

Thus, molar mass of glucose = 180 g mol^-1

Thus, moles of 10 g of glucose `=10/180= 0.0555 moles

Now, Molar mass of water (H₂O)

= 1 × 2 + 16

Thus, molar mass of water = 18 g mol^-1

Thus, moles of 90g of water `=90/18` = 5 moles

Calculation of Molality

Now, Molality of a solution = Moles of solutes/Mass of solvent in kg

Or, Molality of a solution = Moles of solutes × 1000/Mass of solvent (g)

Thus, Molality of the given solution `=(0.0555xx1000)/90`

Thus, Molality of the given solution =0.617 m

Calculation of Mole fraction

We know that, Mole fraction of components = Number of moles of the component/Total number of moles of all the components

And for binary solution, Mole fraction of A(x_A) `=n_A/(n_A+n_B)`

Where A and Ba are components and n_A and n_B are moles respectively.

Thus, Mole fraction of glucose (X_g) = No. of moles of glucose/(No. of moles of glucose+No. of moles of water)

`=>X_g = 0.0555/(5+0.0555) =0.01`

Thus, mole fraction of glucose (X_g) = 0.01

And Mole fraction of water (X_w) `=n_w/(n_g+n_w)`

`=5/(5+0.0555)=0.99`

Thus, mole fraction of water (X_w) = 0.99

Calculation of Molarity

Now, we know that volume of a solution = mass/density

Thus, volume of 100g solution `= 100/1.2`

= 83.33 mL

Thus, volume of given solution = 83.33 mL

Now, We know that, Molarity of a solution = Moles of solute/Volume of solution in litre

Thus, Molarity of the given solution `=(0.0555xx1000)/83.33` M

= 0.67 M

Thus, Answer =

Molality of the given solution =0.617 m

Mole fraction of glucose = 0.01

Mole fraction of water = 0.99

And, Molarity of the given solution = 0.67 M

Question (6) How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer:

Given, Amount of mixture of Na₂CO₃ and NaHCO₃ = 1g

Mixture contains equimolar amounts of both components.

Thus, no of mL of 0.1 M of HCl = ?

Calculation of number of moles

Let amount of Na²CO₃ in mixture = x g

Thus, amount of NaHCO₃ in mixture = (1–x) g

Now, Molar mass of Na²CO₃

= 23 × 2 + 12 + 16 × 3 mol^–1

= 106 mol^–1

And molar mass of NaHCO₃

= 23 × 1 + 1 + 12 + 16 × 3 mol^–1

= 84 mol^–1

Now, number of moles in x g of Na₂CO₃

`=x/106`

And number of moles of NaHCO₃ in (1–x) g

`=(1-x)/84`

Now, according to question, since mixture contains equimolar amount of both of the components

Thus, `x/106=(1-x)/84`

`=>84x=106(1-x)`

`=>84x = 106-106x`

`=>84x+106x=106`

`=>190x=106`

`=>x=106/190`

⇒ x = 0.5574 g

Thus, amount of Na₂CO₃ = 0.5574 g

And, amount of NaHCO₃ `=(1-x)`

= 1 – 0.5574 g

= 0.4426 g

Thus, amount of NaHCO₃ = 0.4426 g

Now, number of moles of Na₂CO₃ present in the given mixture

`=0.558/106` = 0.00526

And number of moles of NaHCO₃ present in the given mixture

`=0.4426/84`

= 0.00526

Calculation of number of moles of HCl required in question

The reaction between sodium carbonate and hydrochloric acid can be written as

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

After observation of reaction, it is clear that each mole of Na₂CO₃ requires two moles of HCl

Now, ∵ 1 mole of Na₂CO₃ requires 2 moles of HCl

∴ 0.00526 moles of Na₂CO₃ requires 0.00526 × 2 moles of HCl

= 0.01053 moles of HCl

And, since 1 mole of NaHCO₃ requires 1 mole of HCl

Thus, 0.00526 moles of NaHCO₃ require 0.00526 moles of HCl

Therefore, total amount of HCl required

= Number of moles of HCl required by Na₂CO₃ + Moles required by NaHCO₃

= 0.01053 + 0.00526

= 0.01579 moles of HCl

Now, according to question,

∵ 0.1 mole of 0.1 M HCl are present in 1000 mL of HCl

∴ 1 mole of 0.1M HCl is present in `1000/0.1` mL of HCl

∴ 0.01579 mole of 0.1 M HCl present in `1000/0.1xx0.1579` mL of HCl

= 157.9 mL of HCl

Thus, Answer = 157.9 mL

Question (7) A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Answer:

Given,

% of first component of solution = 25% solution

And, mass of first component = 300g

And, % of second component = 40%

And, mass of first component = 400 g

Thus, mass percentage of resulting solution = ?

Mass of solute in 300 g of 25% solution = 300 × 25%

`=300/100xx25` = 75 g

And, mass of solute in 400 g of 40% solution = 400 × 40%

`=400/100xx40` = 160 g

Thus, total mass of solute in given mixture = 75 + 160 = 235 g

And, total mass of mixture = 300 g + 400 g = 700 g

Now, we know that, mass percentage of component (w/w)

= Mass of the component in solution×100/Total mass of solution

Thus, mass percentage of solute in given solution `=235/700xx100`

Thus, mass % of solute = 33.57 % Answer

Question (8) An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?

Answer:

Mass of ethylene glycol (C₂H₆O₂) in mixture = 222.6 g

And, mass of water in mixture = 200 g

And, density of the solution = 1.072 mL^–1

Thus, molality and molarity of the solution = ?

Thus, total mass of solution = mass of solute + mass of solvent

= 222.6 g + 200 g

Thus, total mass of solute = 422.6 g

And, Molar mass of ethylene glycol C₂H₆O₂

=2 × 12 + 6 × 1 + 2 × 16

=24+6+32

Thus, molar mass of ethylene glycol = 62 g mol^-1

Now, number of moles of solute (ethylene glycol)

`=(222.6)/(62)=3.6` M

Thus, number of moles of solute (ethylene glycol) = 3.6 M

Calculation of Molality

We know that Molality = Mass of solute/Mass of solvent in kg

Thus, molality of given solution `=3.6/200xx1000`

= 17.95 mol/kg

Calculation of Molarity

Here, total mass of solution = 422.6 g

And density of solution = 1.072 g mL^–1

Thus, volume of solution = Total mass of solution/Density of solution

`=422.6/1072`

= 934.21 mL

Now, we know that, Molarity = Moles of solue/Volume of solution in litre

Thus, molarity of given solution `=3.59/394.21xx1000`

= 9.1 mol L^–1

Thus, molality = 17.95 mol/kg

And, molarity = 9.1 mol _–1 Answer

Question (9) A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃ supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).

(a) Express this in percent by mass

(b) determine the molality of chloroform in the water sample

Solution

Given, the level of contamination = 15 ppm (by mass)

Thus, percent by mass = ?

And, molality of chloroform in water = ?

Here the meaning of 15 ppm is concentration of chloroform is 15 parts per million

Thus, percentage of chloroform in water `=15/10^6xx100`

= 15 × 10^–4%

And mass of solvent = 10⁶ g

[Since, 15 g of chloroform present in 10⁶ of the solution]

Now, molar mass of chloroform (CHCl₃)

= 12 + 1 + 35.5 × 3 mol^–1

Thus, molar mass of chloroform = 119.5 mol^–1

Now, moles of CHCl₃ `=15/119.5`

= 0.1255

Now, we know that Molality = Moles of solute/Mass of solvent in kg

Thus, molality of given solution `=0.1255xx1000/10^6`

= 1.25 × 10^–4 m Answer

Thus, Percentage by mass = 15 × 10^–4%

And, molality = 1.25 × 10^–4 m Answer

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