CBSE Class 12 Chemistry Chapter 2: Solutions - NCERT Solutions

Solution:

To calculate the mass percentage of each component in the solution, we first find the total mass of the solution.

Mass of benzene = 22 g

Mass of carbon tetrachloride = 122 g

Total mass of the solution = Mass of benzene + Mass of carbon tetrachloride

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

Now, we calculate the mass percentage for each component:

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

$$=\; \backslash frac\{22\; \backslash text\{\; g\}\}\{144\; \backslash text\{\; g\}\}\; \backslash times\; 100\backslash \%\; \backslash approx\; 15.28\backslash \%$$

Mass percentage of carbon tetrachloride = $$\backslash frac\{\backslash text\{Mass\; of\; carbon\; tetrachloride\}\}\{\backslash text\{Total\; mass\; of\; the\; solution\}\}\; \backslash times\; 100\backslash \%$$

$$=\; \backslash frac\{122\; \backslash text\{\; g\}\}\{144\; \backslash text\{\; g\}\}\; \backslash times\; 100\backslash \%\; \backslash approx\; 84.72\backslash \%$$

Alternatively, the mass percentage of carbon tetrachloride can be found by subtracting the mass percentage of benzene from 100%:

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

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

Solution:

We are given that the solution contains 30% benzene by mass. Let's assume the total mass of the solution is 100 g. This means:

Mass of benzene = 30 g

Mass of carbon tetrachloride (CCl₄) = 100 g - 30 g = 70 g

Next, we need to find the molar masses of benzene and carbon tetrachloride to calculate their respective moles.

Molar mass of benzene (C₆H₆) = (6 × Atomic mass of C) + (6 × Atomic mass of H)

$$=\; (6\; \backslash times\; 12.011\; \backslash text\{\; g\; mol\}^\{-1\})\; +\; (6\; \backslash times\; 1.008\; \backslash text\{\; g\; mol\}^\{-1\})\; \backslash approx\; 78.11\; \backslash text\{\; g\; mol\}^\{-1\}$$

Number of moles of benzene = $$\backslash frac\{\backslash text\{Mass\; of\; benzene\}\}\{\backslash text\{Molar\; mass\; of\; benzene\}\}\; =\; \backslash frac\{30\; \backslash text\{\; g\}\}\{78.11\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.384\; \backslash text\{\; mol\}$$

Molar mass of carbon tetrachloride (CCl₄) = (1 × Atomic mass of C) + (4 × Atomic mass of Cl)

$$=\; (1\; \backslash times\; 12.011\; \backslash text\{\; g\; mol\}^\{-1\})\; +\; (4\; \backslash times\; 35.45\; \backslash text\{\; g\; mol\}^\{-1\})\; \backslash approx\; 12.011\; +\; 141.8\; \backslash text\{\; g\; mol\}^\{-1\}\; \backslash approx\; 153.81\; \backslash text\{\; g\; mol\}^\{-1\}$$

Number of moles of carbon tetrachloride = $$\backslash frac\{\backslash text\{Mass\; of\; CCl\}\_4\}\{\backslash text\{Molar\; mass\; of\; CCl\}\_4\}\; =\; \backslash frac\{70\; \backslash text\{\; g\}\}\{153.81\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.455\; \backslash text\{\; mol\}$$

The mole fraction of benzene is calculated as:

Mole fraction of benzene = $$\backslash frac\{\backslash text\{Number\; of\; moles\; of\; benzene\}\}\{\backslash text\{Total\; number\; of\; moles\; (benzene\; +\; CCl\}\_4\backslash text\{)\}\}$$

$$=\; \backslash frac\{0.384\; \backslash text\{\; mol\}\}\{0.384\; \backslash text\{\; mol\}\; +\; 0.455\; \backslash text\{\; mol\}\}\; =\; \backslash frac\{0.384\}\{0.839\}\; \backslash approx\; 0.458$$

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

Solution:

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

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

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

First, calculate the molar mass of cobalt(II) nitrate hexahydrate (Co(NO₃)₂. 6H₂O).

Atomic masses: Co = 59, N = 14, O = 16, H = 1

Molar mass of Co(NO₃)₂. 6H₂O = (Atomic mass of Co) + 2 × (Atomic mass of N + 3 × Atomic mass of O) + 6 × (2 × Atomic mass of H + Atomic mass of O)

$$=\; 59\; +\; 2\; \backslash times\; (14\; +\; 3\; \backslash times\; 16)\; +\; 6\; \backslash times\; (2\; \backslash times\; 1\; +\; 16)$$

$$=\; 59\; +\; 2\; \backslash times\; (14\; +\; 48)\; +\; 6\; \backslash times\; (2\; +\; 16)$$

$$=\; 59\; +\; 2\; \backslash times\; 62\; +\; 6\; \backslash times\; 18$$

$$=\; 59\; +\; 124\; +\; 108\; =\; 291\; \backslash text\{\; g\; mol\}^\{-1\}$$

Now, calculate the number of moles of the solute:

Number of moles = $$\backslash frac\{\backslash text\{Mass\; of\; solute\}\}\{\backslash text\{Molar\; mass\; of\; solute\}\}\; =\; \backslash frac\{30\; \backslash text\{\; g\}\}\{291\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.103\; \backslash text\{\; mol\}$$

The volume of the solution is given as 4.3 L.

Calculate the molarity:

Molarity = $$\backslash frac\{0.103\; \backslash text\{\; mol\}\}\{4.3\; \backslash text\{\; L\}\}\; \backslash approx\; 0.0239\; \backslash text\{\; M\}$$

Rounding to two significant figures, the molarity is 0.024 M.

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

We 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

Substitute the values into the formula:

$$0.5\; \backslash text\{\; M\}\; \backslash times\; 30\; \backslash text\{\; mL\}\; =\; M\_2\; \backslash times\; 500\; \backslash text\{\; mL\}$$

Solve for M₂:

$$M\_2\; =\; \backslash frac\{0.5\; \backslash text\{\; M\}\; \backslash times\; 30\; \backslash text\{\; mL\}\}\{500\; \backslash text\{\; mL\}\}\; =\; \backslash frac\{15\}\{500\}\; \backslash text\{\; M\}\; =\; 0.03\; \backslash text\{\; M\}$$

Answer:

(a) The molarity of the Co(NO₃)₂. 6H₂O solution is approximately 0.024 M.

(b) The molarity of the diluted H₂SO₄ solution is 0.03 M.