CBSE Class 12 Chemistry Chapter 2: Solutions - NCERT Solutions

To calculate the mass percentage of each component in the solution, we use the formula:

$$Mass\; \backslash \%\; of\; component\; =\; \backslash frac\{\backslash text\{Mass\; of\; component\}\}\{\backslash text\{Total\; mass\; of\; solution\}\}\; \backslash times\; 100\backslash \%$$

Given:

• Mass of benzene (C₆H₆) = 22 g
• Mass of carbon tetrachloride (CCl₄) = 122 g

First, calculate the total mass of the solution:

$$Total\; mass\; of\; solution\; =\; Mass\; of\; benzene\; +\; Mass\; of\; CCl\_4$$

$$Total\; mass\; of\; solution\; =\; 22\; \backslash ,\; \backslash text\{g\}\; +\; 122\; \backslash ,\; \backslash text\{g\}\; =\; 144\; \backslash ,\; \backslash text\{g\}$$

Now, calculate the mass percentage of benzene:

$$Mass\; \backslash \%\; of\; C\_6H\_6\; =\; \backslash frac\{22\; \backslash ,\; \backslash text\{g\}\}\{144\; \backslash ,\; \backslash text\{g\}\}\; \backslash times\; 100\backslash \%$$

$$Mass\; \backslash \%\; of\; C\_6H\_6\; \backslash approx\; 15.28\backslash \%$$

Next, calculate the mass percentage of carbon tetrachloride:

$$Mass\; \backslash \%\; of\; CCl\_4\; =\; \backslash frac\{122\; \backslash ,\; \backslash text\{g\}\}\{144\; \backslash ,\; \backslash text\{g\}\}\; \backslash times\; 100\backslash \%$$

$$Mass\; \backslash \%\; of\; CCl\_4\; \backslash approx\; 84.72\backslash \%$$

Alternatively, since there are only two components, the mass percentage of CCl₄ can be calculated as:

$$Mass\; \backslash \%\; of\; CCl\_4\; =\; (100\backslash \%\; -\; \backslash text\{Mass\; \backslash \%\; of\; C\}\_6\backslash text\{H\}\_6)$$

$$Mass\; \backslash \%\; of\; CCl\_4\; =\; (100\backslash \%\; -\; 15.28\backslash \%)\; =\; 84.72\backslash \%$$

Answer: The mass percentage of benzene is approximately 15.28% and the mass percentage of carbon tetrachloride is approximately 84.72%.

We are given that the solution contains 30% benzene by mass. This means that in a given mass of the solution, 30% of it is benzene and the remaining is carbon tetrachloride.

Let's assume the total mass of the solution is 100 g for easier calculation.

• Mass of benzene (C₆H₆) = 30% of 100 g = 30 g
• Mass of carbon tetrachloride (CCl₄) = 100 g - 30 g = 70 g

To calculate the mole fraction, we first need to find the number of moles of each component. We need their molar masses:

• Molar mass of benzene (C₆H₆) = (6 × Atomic mass of C) + (6 × Atomic mass of H)
• Molar mass of C₆H₆ = (6 × 12.011 g/mol) + (6 × 1.008 g/mol) ≈ 78.11 g/mol
• Molar mass of carbon tetrachloride (CCl₄) = (1 × Atomic mass of C) + (4 × Atomic mass of Cl)
• Molar mass of CCl₄ = (1 × 12.011 g/mol) + (4 × 35.45 g/mol) ≈ 153.81 g/mol

Now, calculate the number of moles for each component:

$$Number\; of\; moles\; of\; C\_6H\_6\; =\; \backslash frac\{\backslash text\{Mass\; of\; C\}\_6\backslash text\{H\}\_6\}\{\backslash text\{Molar\; mass\; of\; C\}\_6\backslash text\{H\}\_6\}$$

$$Number\; of\; moles\; of\; C\_6H\_6\; =\; \backslash frac\{30\; \backslash ,\; \backslash text\{g\}\}\{78.11\; \backslash ,\; \backslash text\{g/mol\}\}\; \backslash approx\; 0.384\; \backslash ,\; \backslash text\{mol\}$$

$$Number\; of\; moles\; of\; CCl\_4\; =\; \backslash frac\{\backslash text\{Mass\; of\; CCl\}\_4\}\{\backslash text\{Molar\; mass\; of\; CCl\}\_4\}$$

$$Number\; of\; moles\; of\; CCl\_4\; =\; \backslash frac\{70\; \backslash ,\; \backslash text\{g\}\}\{153.81\; \backslash ,\; \backslash text\{g/mol\}\}\; \backslash approx\; 0.455\; \backslash ,\; \backslash text\{mol\}$$

The mole fraction of benzene (X_C6H6) is calculated as:

$$X\_\{\backslash text\{C\}\_6\backslash text\{H\}\_6\}\; =\; \backslash frac\{\backslash text\{Number\; of\; moles\; of\; C\}\_6\backslash text\{H\}\_6\}\{\backslash text\{Number\; of\; moles\; of\; C\}\_6\backslash text\{H\}\_6\; +\; \backslash text\{Number\; of\; moles\; of\; CCl\}\_4\}$$

$$X\_\{\backslash text\{C\}\_6\backslash text\{H\}\_6\}\; =\; \backslash frac\{0.384\; \backslash ,\; \backslash text\{mol\}\}\{0.384\; \backslash ,\; \backslash text\{mol\}\; +\; 0.455\; \backslash ,\; \backslash text\{mol\}\}$$

$$X\_\{\backslash text\{C\}\_6\backslash text\{H\}\_6\}\; =\; \backslash frac\{0.384\}\{0.839\}\; \backslash approx\; 0.458$$

Answer: The mole fraction of benzene in the solution is approximately 0.458.

Molarity (M) is defined as the number of moles of solute dissolved per liter of solution. The formula is:

$$Molarity\; (M)\; =\; \backslash frac\{\backslash text\{Moles\; of\; solute\}\}\{\backslash text\{Volume\; of\; solution\; in\; Liters\}\}$$

(a) Molarity of 30 g of Co(NO₃)₂. 6H₂O in 4.3 L of solution

First, we need to calculate the molar mass of the solute, Cobalt(II) nitrate hexahydrate (Co(NO₃)₂. 6H₂O).

• Atomic mass of Co = 59 g/mol
• Atomic mass of N = 14 g/mol
• Atomic mass of O = 16 g/mol
• Atomic mass of H = 1 g/mol

Molar mass of Co(NO₃)₂. 6H₂O = (59) + 2 × (14 + 3 × 16) + 6 × (2 × 1 + 16)

$$Molar\; mass\; =\; 59\; +\; 2\; \backslash times\; (14\; +\; 48)\; +\; 6\; \backslash times\; (18)$$

$$Molar\; mass\; =\; 59\; +\; 2\; \backslash times\; (62)\; +\; 6\; \backslash times\; 18$$

$$Molar\; mass\; =\; 59\; +\; 124\; +\; 108\; =\; 291\; \backslash ,\; \backslash text\{g/mol\}$$

Next, calculate the number of moles of the solute:

$$Moles\; of\; solute\; =\; \backslash frac\{\backslash text\{Mass\; of\; solute\}\}\{\backslash text\{Molar\; mass\; of\; solute\}\}$$

$$Moles\; of\; solute\; =\; \backslash frac\{30\; \backslash ,\; \backslash text\{g\}\}\{291\; \backslash ,\; \backslash text\{g/mol\}\}\; \backslash approx\; 0.103\; \backslash ,\; \backslash text\{mol\}$$

The volume of the solution is given as 4.3 L.

Now, calculate the molarity:

$$Molarity\; =\; \backslash frac\{0.103\; \backslash ,\; \backslash text\{mol\}\}\{4.3\; \backslash ,\; \backslash text\{L\}\}\; \backslash approx\; 0.024\; \backslash ,\; \backslash text\{M\}$$

Answer (a): The molarity of the solution is approximately 0.024 M.

(b) Molarity of 30 mL of 0.5 M H₂SO₄ diluted to 500 mL

We can use the dilution formula, M₁V₁ = M₂V₂, where:

• M₁ = Initial molarity = 0.5 M
• V₁ = Initial volume = 30 mL
• M₂ = Final molarity (what we need to find)
• V₂ = Final volume = 500 mL

Rearranging the formula to solve for M₂:

$$M\_2\; =\; \backslash frac\{M\_1\; V\_1\}\{V\_2\}$$

Substitute the given values:

$$M\_2\; =\; \backslash frac\{(0.5\; \backslash ,\; \backslash text\{M\})\; \backslash times\; (30\; \backslash ,\; \backslash text\{mL\})\}\{500\; \backslash ,\; \backslash text\{mL\}\}$$

$$M\_2\; =\; \backslash frac\{15\}\{500\}\; \backslash ,\; \backslash text\{M\}$$

$$M\_2\; =\; 0.03\; \backslash ,\; \backslash text\{M\}$$

Answer (b): The molarity of the diluted solution is 0.03 M.