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#### Example – 01:

- A solution of NaOH (molar mass 40 g mol
^{-1}) was prepared by dissolving 1.6 g of NaOH in 500 cm^{3}of water. Calculate molarity of the NaOH solution. **Given:**Mass of NaOH = 1.6 g, molar mass of NaOH = 40 g mol^{-1}, volume of water = 500 cm^{3}= 500 mL = 0.5 L**To Find:**Molarity =?**Solution:**

Number of moles = Given mass/ Molecular mass = 1.6 g/40 g mol^{-1 } = 0.04 mol

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

Molarity = 0.04 mol /0.5 L = 0.08 mol L^{-1}

**Ans:** The molarity of NaOH solution is 0.08 mol L^{-1 }or 0.08 M

#### Example – 02:

- Calculate molarity of 4 g caustic soda dissolved in 200 mL of solution.
**Given:**Mass of solute (caustic soda) = 4 g, volume of solution = 200 mL = 0.2 L**To Find:**Molarity of the solution = ?**Solution:**

Molar mass of caustic solute (caustic soda NaOH) = 23 g x1 + 16 g x 1 + 1 g x 1 = (23 + 16 + 1) g = 40 g

Number of moles of caustic solute (caustic soda) = given mass/molecular mass = 4 g/ 40 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

**Ans:** The molarity of caustic soa solution is 0.5 mol L^{-1 }or 0.5 M

#### Example – 03:

- Calculate molarity of 5.3 g anhydrous sodium carbonate dissolved in 100 mL of solution.
**Given:**Mass of solute (sodium carbonate) = 5.3 g, volume of solution = 100 mL = 0.1 L**To Find:**Molarity of the solution = ?**Solution:**

Molar mass of (Na_{2}CO_{3}) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of (Na_{2}CO_{3}) = given mass/molecular mass = 5.3 g/ 106 g = 0.05

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.05/0.1 = 0.5 M

**Ans:** The molarity of sodium carbonate solution is 0.5 mol L^{-1 }or 0.5 M

#### Example – 04:

- Calculate molarity of 0.365 g pure HCl gas dissolved in 50 mL of solution.
**Given:**Mass of solute (HCl) = 0.365 g, volume of solution = 50 mL = 0.05 L**To Find:**Molarity of the solution = ?**Solution:**

Molar mass of HCl = 1 g x 1 + 35.5 g x 1 = (1 + 35.5) g = 36.5 g

Number of moles of HCl = given mass/molecular mass = 0.365 g/ 36.5 g = 0.01

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.01/0.05 = 0.2 M

**Ans:** The molarity of HCl solution is 0.2 mol L^{-1 }or 0.2 M

#### Example – 05:

- Calculate molarity of 5.85 g NaCl dissolved in 200 mL of solution.
**Given:**Mass of solute (NaCl) = 5.85 g, volume of solution = 200 mL = 0.2 L**To Find:**Molarity of the solution = ?**Solution:**

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 5.85 g/ 58.5 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

**Ans:** The molarity of NaCl solution is 0.5 mol L^{-1 }or 0.5 M

#### Example – 06:

- Calculate molarity of 20.6 g NaBr dissolved in 500 mL of solution.
**Given:**Mass of solute (NaCl) = 20.6 g, volume of solution = 500 mL = 0.5 L**To Find:**Molarity of the solution = ?**Solution:**

Molar mass of NaBr = 23 g x 1 + 80 g x 1 = (23 + 80) g = 103 g

Number of moles of NaBr = given mass/molecular mass = 20.6 g/ 103 g = 0.2

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2/0.5 = 0.4 M

**Ans:** The molarity of NaBr solution is 0.4 mol L^{-1 }or 0.4 M

#### Example – 07:

- Calculate molarity of pure water if its density is 1 g/mL.
**Given:**Density of water = 1 g/mL**To Find:**Molarity of pure water =?**Solution:**

Let us consider 1000 mL of water

Mass of water = volume x density = 1000 mL x 1 g/mL = 1000 g

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

Number of moles of water = given mass/molecular mass = 1000/ 18 g = 55.5

Molarity of pure water = Number of moles of the solute/volume of solution in L = 55.55/1 = 55.55 M

**Ans:** The molarity of pure water is 55.55 mol L^{-1 }or 55.5 M

#### Example – 08:

- Calculate the quantity of anhydrous sodium carbonate required to produce 250 mL decimolar solution.
**Given:**volume of solution = 250 mL = 0.25 L, molarity = decimolar = M/10 = 0.1 M**To Find:**Mass of anhydrous sodium carbonate =?**Solution:**

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

0.1 = Number of moles of the solute/0.25

Number of moles of the solute = 0.1 x 0.25 = 0.025 mol

Molar mass of (Na_{2}CO_{3}) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of = given mass/molecular mass

Mass of Na_{2}CO_{3 }= Number of moles x molecular mass

Mass of Na_{2}CO_{3 }= 106 x 0.025 = 2.65 g

**Ans:** The quantity of sodium carbonate required is 2.65 g

#### Example – 09:

- Sulphuric acid is 95.8 % by mass. Calculate molarity and mole fraction of H
_{2}SO_{4}of density 1.91 g cm^{-3}. Given H = 1, S = 32, O = 16. **Given:**% by mass = 95.8 %, Density of solution = 1.91 g cm^{-3}**To Find:**Mole fraction = ? molarity = ?**Solution:**

Consider 100 g of solution

Mass of H_{2}SO_{4} = 95.8 g and mass of H_{2}O = 100 – 95.8 g = 4.2 g

Molecular mass of water (H_{2}O) = 1 g x 2 + 16 g x 1 = 18 g mol^{-1}

Molecular mass H_{2}SO_{4} = 1 g x 2 + 32 g x 1 + 16g x 4 = 98 g mol^{-1}

Number of moles of water = n_{A} = 4.2 g/ 18 g = 0.2333 mol

Number of moles of H_{2}SO_{4} = n_{B} = 95.8 g/ 98 g = 0.9776 mol

Total number of moles = n_{A} + n_{B} + n_{C} = 0.2333 + 0.9776 = 1.2109

Mole fraction of H_{2}SO_{4} = x_{B} = n_{B}/(n_{A }+n_{B}) = 0.9776/1.2109 = 0.8073

Density of solution = 1.91 g cm^{-3}

Volume of solution = Mass of solution / density = 100 g /1.91 g cm^{-3} = 52.36 cm^{3} = 52.36 mL = 0.05236 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.9776/0.05236 = 18.67 M

**Ans:** The mole fraction of H_{2}SO_{4} is 0.8073 and molarity of solution is 18.67 mol L^{-1 }or 18.67 M

#### Example – 10:

- Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm
^{-3}, calculate molarity of HCl solution and also calculate the mole fraction of HCl and H_{2}O. **Given:**% by mass = 38 %, Density of solution = 1.1 g cm^{-3}**To Find:**Mole fraction = ? molarity = ?**Solution:**

Consider 100 g of solution

Mass of HCl = 38 g and mass of H_{2}O = 100 – 38 g = 62 g

Molecular mass of water (H_{2}O) = 1 g x 2 + 16 g x 1 = 18 g mol^{-1}

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^{-1}

Number of moles of water = n_{A} = 62 g/ 18 g = 3.444 mol

Number of moles of HCl = n_{B} = 38 g/ 36.5 g = 1.041 mol

Total number of moles = n_{A} + n_{B} + n_{C} = 3.444 + 1.041 = 4.485

Mole fraction of HCl = x_{B} = n_{B}/(n_{A }+n_{B}) = 1.041/4.485 = 0.2321

Mole fraction of H_{2}O = x_{A} = n_{A}/(n_{A }+n_{B}) = 3.444/4.485 = 0.7679

Density of solution = 1.1 g cm^{-3}

Volume of solution = Mass of solution / density = 100 g /1.1 g cm^{-3} = 90.91 cm^{3} = 90.91 mL = 0.09091 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.041/0.09091 = 11.45 M

**Ans:** Molarity of solution is 11.45 mol L^{-1 }or 11.45 M

The molefraction of HCl is 0.2321 and that of H_{2}O is 0.7679

#### Example – 11:

- Commercially available concentrated hydrochloric acid is an aqueous solution containing 40% HCl gas by mass. If its density is 1.2 g cm
^{-3}, calculate molarity of HCl solution. **Given:**% by mass = 40 %, Density of solution = 1.2 g cm^{-3}**To Find:**Molarity = ?**Solution:**

Consider 100 g of solution

Mass of HCl = 40 g and mass of H_{2}O = 100 – 40 g = 60 g

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^{-1}

Number of moles of HCl = n_{B} = 40 g/ 36.5 g = 1.096 mol

Density of solution = 1.2 g cm^{-3}

Volume of solution = Mass of solution / density = 100 g /1.2 g cm^{-3} = 83.33 cm^{3} = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.096/0.08333 = 13.15 M

**Ans:** Molarity of solution is 13.15 mol L^{-1 }or 13.15 M

#### Example – 12:

- Calculate molarity of a solution containing 50 g of NaCl in 500 g of solution and having density 0.936 g/cm
^{3}. - Given: Mass of solute (NaCl) = 50 g, mass of solution = 500 g, density of solution = d = 0.936 g/cm
^{3}. - To Find: Molarity of solution = M = ?

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 50 g/ 58.5 g = 0.8547

Density of solution = 0.936 g cm^{-3}

Volume of solution = Mass of solution / density = 500 g /0.936 g cm^{-3} = 534.2 cm^{3} = 534.2 mL = 0.5342 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.8547/0.5342 = 1.6 M

**Ans:** The molarity of NaCl solution is 1.6 mol L^{-1 }or 1.6 M

Science > Chemistry > Solutions and Colligative Properties > You are Here |

Physics |
Chemistry |
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Mathematics |