74.5 g of kcl was dissolved in 1000. ml of water. what is the molality of the solution

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In this article, we shall study numerical problems to calculate molality of a solution.

Example – 01:

7.45 g of potassium chloride (KCl) was dissolved in 100 g of water. Calculate the molality of the solution.

Given: mass of solute (KCl) = 7.45 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 7.45 g/ 74.5 g mol-1 = 0.1 mol

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

Molality = 0.1 mol /0.1 kg = 1 mol kg-1

Ans: The molality of solution is 1 mol kg-1 or 1 m.

Example – 02:

11.11 g of urea (NH2CONH2) was dissolved in 100 g of water. Calculate the molarity and molality of the solution. Given N = 14, H = 1, C = 12, O = 16.

Given: mass of solute (urea) = 11.11 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of urea (NH2CONH2) = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1

Molecular mass of urea (NH2CONH2) = 60 g mol-1

Number of moles of solute (urea) = given mass/ molecular mass

Number of moles of solute (urea) = 11.11 g/ 60 g mol-1 = 0.1852 mol

Volume of water = mass of water/ density = 100 g/1 g mL-1 = 100 mL = 0.1 L

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

Molarity = 0.1852 mol /0.1 L = 1.852 mol L-1 or 1.852 mol dm-3

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

Molality = 0.1852 mol /0.1 kg = 1.852 mol kg-1

Ans: The molarity of solution is 1.852 mol L-1 and the molality is 1.852 mol kg-1

Example – 03:

34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16.

Given: Mass of solute (sugar) = 34.2 g, Mass of solution (sugar syrup) = 214.2 g

To Find: Molality and mole fraction =?

Solution:

Mass of Solution = Mass of solute + mass of solvent

Mass of solvent = mass of solution – mass of solute = 214.2 g – 34.2 g = 180 g = 0.180 kg

Molar mass of sugar (C12H22O11) = 12 g x 12 + 1 g x 22 + 16 g x 11 = 342 g mol-1

Number of moles of solute (sugar) = nB = Given mass/ molecular mass = 34.2 g/342 g mol-1  = 0.1 mol

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

Molality = 0.1 mol /0.180 kg = 0.5556 mol kg-1

Molar mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Number of moles of solvent (water) = nA = Given mass/ molecular mass = 180 g/18 g mol-1  = 10 mol

Total number of moles = nA + nB = 0.1 + 10 = 10.1 mol

Mole fraction of solute (sugarl) = xB = nB/(nA + nB) = 0.1/10.1 = 0.0099

Mole fraction of sugar = 0.0099

Ans: Molality of solution = 0.5556 mol kg-1 and mole fraction of sugar = 0.0099

Example – 04:

10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm-3, calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol-1, Cl = 35.5 g mol-1.

Given: the mass of solute (KCl) = 10 g, the mass of solvent (water) = 1000 g = 1 kg, density of solution = 0.997 g cm-3,

To Find: molarity =? molality = ?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 10 g/ 74.5 g mol-1 = 0.1342 mol

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

Molality = 0.1342 mol /1 kg = 0.1342 mol kg-1

Mass of solution = 10 g + 1000 g = 1010 g

Volume of solution = mass of solution/density = 1010/0.997 g cm-3

Volume of solution = 1013 cm3 = 1013 mL = 1.013 L

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

Molarity = 0.1342 mol /1.013 L = 0.1325 mol L-1

Ans: The molarity of the solution is 0.1325 mol L-1 or 0.1325 M, the molality of the solution is 0.1342 mol kg-1 or 0.1342 m.

Example – 05:

Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm-3 containing 27 % by mass of sulphuric acid.

Given: density of the solution = 1.198 g cm-3, % mass of sulphuric acid = 27%,

To Find: Molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of H2SO4 = 27 g and mass of H2O = 100 – 27 g = 73 g = 0.073 kg

Molecular mass H2SO4 = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol-1

Number of moles of H2SO4 = nB = 27 g/ 98 g = 0.2755 mol

Density of solution = 1.198 g cm-3

Volume of solution = Mass of solution / density = 100 g /1.198 g cm-3 = 83.47 cm3 = 83.47 mL = 0.08347 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2755/0.08347 = 3.301 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.2755 mol /0.073 kg = 3.774 mol L-1

Ans: The molarity of solution is 3.374 mol L-1 or 3.374 M, the molality of solution is 3.774 mol L-1 or 3.774 m

Example – 06:

Calculate the mole fraction, molality and molarity of HNO3 in a solution containing 12.2 % HNO3. Given density of HNO3 as 1.038 g cm-3, H = 1, N = 14, O = 16.

Given: density of the solution = 1.038 g cm-3, % mass of HNO3 = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO3 = 12.2 g and mass of H2O = 100 – 12.2 g = 87.8 g = 0.0878 kg

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molecular mass HNO3 = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol-1

Number of moles of water = nA = 87.8 g/ 18 g = 4.8778 mol

Number of moles of HNO3 = nB = 12.2 g/ 63 g = 0.1937 mol

Total number of moles = nA + nB + nC = 4.8778 + 0.1937 = 5.0715

Mole fraction of HNO3 = xB = nB/(nA +nB) = 0.1937/5.0715 = 0.0382

Density of solution = 1.038 g cm-3

Volume of solution = Mass of solution / density = 100 g /1.038 g cm-3 = 96.34 cm3 = 96.34 mL = 0.09634 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1937/0.09634 =2.011 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1937 mol /0.0878 kg = 2.206 mol kg-1

Ans: The mole fraction of HNO3 is 0. 0382, the molarity of solution is 2.011 mol L-1 or 2.011 M, the molality of solution is 2.206 mol kg-1 or 2.206 m

Example – 07:

Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm-3. Given atomic masses H = 1, N = 14 and O = 16.

Given: density of the solution = 1.04 g cm-3, % mass of HNO3 = 6.3 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO3 = 6.3 g and mass of H2O = 100 – 6.3 g = 93.7 g = 0.0937 kg

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molecular mass HNO3 = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol-1

Number of moles of water = nA = 93.4 g/ 18 g = 5.189 mol

Number of moles of HNO3 = nB = 6.3 g/ 63 g = 0.1 mol

Density of solution = 1.04 g cm-3

Volume of solution = Mass of solution / density = 100 g /1.04 g cm-3 = 96.15 cm3 = 96.15 mL = 0.09615 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.09615 =1.040 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1 mol /0.0937 kg = 1.067 mol kg-1

Ans: The molarity of solution is 1.040 mol L-1 or 1.040 M

The molality of solution is 1.067 mol kg-1 or 1.067 m

Example – 08:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm-3. Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Given: density of the solution = 1.038 g cm-3, % mass of HNO3 = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of NaOH = 10 g and mass of H2O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molecular mass NaOH = 23 g x 1 + 16 g x 1 + 1 g  x 1 = 40 g mol-1

Number of moles of water = nA = 90 g/ 18 g = 5 mol

Number of moles of NaOH = nB = 10 g/ 40 g = 0.25 mol

Total number of moles = nA + nB = 5 + 0.25 = 5.25 mol

Mole fraction of NaOH = xB = nB/(nA +nB) = 0.25/5.25 = 0.0476

Density of solution = 1.070 g cm-3

Volume of solution = Mass of solution / density = 100 g /1.070 g cm-3 = 93.46 cm3 = 93.46 mL = 0.09346 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.25/0.09346 =2.675 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.25 mol /0.090 kg = 2.778 mol kg-1

Ans: The molarity of solution is 2.675mol L-1 or 2.675 M, the molality of solution is 2.778 mol kg-1 or 2.778 m, the mole fraction of NaOH is 0. 0476

Example – 09:

A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL-1 and molar mass of glucose is 180 g mol-1.

Given: density of the solution = 1.20 g cm-3, % mass of glucose = 10 %, molar mass of glucose is 180 g mol-1.

To Find: molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of glucose = 10 g and mass of H2O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molecular mass glucose = 180 g mol-1

Number of moles of water = nA = 90 g/ 18 g = 5 mol

Number of moles of glucose = nB = 10 g/ 180 g = 0.0556 mol

Density of solution = 1.20 g cm-3

Volume of solution = Mass of solution / density = 100 g /1.20 g cm-3 = 83.33 cm3 = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.0556/0.08333 =0.6672 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.0556 mol /0.090 kg = 0.6178 mol kg-1

Ans: The molarity of solution is 0.6672 mol L-1 or 0.6672 M, the molality of solution is 0.6178 mol kg-1 or 0.6178 m,

Example – 10:

Battery acid 4.22 M aqueous H2SO4 solution, and has density 1.21 g cm-3. What is the molality of H2SO4. Given H = 1, S = 32, O = 16

Given: density of the solution = 1.21 g cm-3, Molarity of solution = 4.22 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H2SO4 = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 4.22 x 1 = 4.22

Density of solution = 1.21 g cm-3 = 1.21 g/mL = 1.21 x 103 g/L = 1.21 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.21 kg/L = 1.21 kg

Mass of solute (H2SO4) = Number of moles x molecular mass = 4.22 x 98

Mass of solute (H2SO4) = 413.56 g = 0.41356 kg

Mass of solvent = mass of solution – mass of solute = 1.21 – 0.41356 = 0.79644 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 4.22 mol /0.79644 kg = 5.298 mol kg-1

Ans: Molality of solution is 5.298 mol kg-1 or 5.298 m

Example – 11:

The density of 5.35 M H2SO4 solution is 1.22 g cm-3. What is molality of a solution?

Given: density of the solution = 1.22 g cm-3, Molarity of solution = 5.35 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H2SO4 = 1 g x 2 + 32 g x 1 + 16g x 4 = 98 g mol-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 5.35 x 1 = 5.35

Density of solution = 1.22 g cm-3 = 1.22 g/mL = 1.22 x 103 g/L = 1.22 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.22 kg/L = 1.22 kg

Mass of solute (H2SO4) = Number of moles x molecular mass = 5.35 x 98

Mass of solute (H2SO4) = 524.3 g = 0.5243 kg

Mass of solvent = mass of solution – mass of solute = 1.22 – 0.5243 = 0.6957 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 5.35 mol /0.6957 kg = 7.690 mol kg-1

Ans: Molality of solution is 7.690 mol kg-1 or 7.690 m

Example – 12:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given: molality = 2 molal

To Find: Mole fraction =?

Solution:

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molality of solution = 2 molal = 2 mol mol kg-1

The number of moles of solute = 2

The mass of solvent (water) = 1 kg = 1000 g

Number of moles of solvent (water) = 1000/16 = 55.55

Mole fraction of solute = 2/(2 + 55.55) = 2/57.55 = 0.03475

Ans: Mole fraction of solute is 0.0345

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