CBSE Class 12 Chemistry Chapter 4: Solutions NCERT Solutions

Solution:

To calculate the mass percentage of each component in the solution, we first need to 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 can calculate the mass percentage for each component:

Mass percentage of benzene:

$$\backslash text\{Mass\; \backslash \%\; of\; \}\; C\_6H\_6\; =\; \backslash frac\{\backslash text\{Mass\; of\; \}\; C\_6H\_6\}\{\backslash text\{Total\; mass\; of\; solution\}\}\; \backslash times\; 100\backslash \%$$

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

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

Mass percentage of carbon tetrachloride:

$$\backslash text\{Mass\; \backslash \%\; of\; \}\; CCl\_4\; =\; \backslash frac\{\backslash text\{Mass\; of\; \}\; CCl\_4\}\{\backslash text\{Total\; mass\; of\; solution\}\}\; \backslash times\; 100\backslash \%$$

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

$\backslash text\{Mass\; \backslash \%\; of\; \}\; CCl\_4\; \backslash approx\; 84.72\backslash \%$

Alternatively, since the sum of mass percentages must be 100%, we can calculate the mass percentage of CCl₄ as:

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

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.

Mass of benzene = 30% of 100 g = 30 g

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

Next, we need to find the number of moles of each component. We'll use their molar masses:

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

$Molar\; mass\; of\; \}\; C\_6H\_6\; =\; (6\; \backslash times\; 12.011\; \backslash text\{\; g\; mol\}^\{-1\})\; +\; (6\; \backslash times\; 1.008\; \backslash text\{\; g\; mol\}^\{-1\})\; \backslash approx\; 78.11\; \backslash text\{\; g\; mol\}^\{-1\}$

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

$n\_\{C\_6H\_6\}\; =\; \backslash frac\{30\; \backslash text\{\; g\}\}\{78.11\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.3841\; \backslash text\{\; mol\}$

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

$Molar\; mass\; of\; \}\; CCl\_4\; =\; (1\; \backslash times\; 12.011\; \backslash text\{\; g\; mol\}^\{-1\})\; +\; (4\; \backslash times\; 35.45\; \backslash text\{\; g\; mol\}^\{-1\})\; \backslash approx\; 12.011\; +\; 141.8\; \backslash text\{\; g\; mol\}^\{-1\}\; \backslash approx\; 153.81\; \backslash text\{\; g\; mol\}^\{-1\}$

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

$n\_\{CCl\_4\}\; =\; \backslash frac\{70\; \backslash text\{\; g\}\}\{153.81\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.4551\; \backslash text\{\; mol\}$

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

$$\backslash chi\_\{C\_6H\_6\}\; =\; \backslash frac\{\backslash text\{Number\; of\; moles\; of\; benzene\}\}\{\backslash text\{Total\; number\; of\; moles\; in\; the\; solution\}\}\; =\; \backslash frac\{n\_\{C\_6H\_6\}\}\{n\_\{C\_6H\_6\}\; +\; n\_\{CCl\_4\}\}$$

$$\backslash chi\_\{C\_6H\_6\}\; =\; \backslash frac\{0.3841\; \backslash text\{\; mol\}\}\{0.3841\; \backslash text\{\; mol\}\; +\; 0.4551\; \backslash text\{\; mol\}\}\; =\; \backslash frac\{0.3841\}\{0.8392\}$$

$\backslash chi\_\{C\_6H\_6\}\; \backslash approx\; 0.4577$

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 per liter of solution.

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

(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 g/mol, N = 14 g/mol, O = 16 g/mol, H = 1 g/mol.

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)

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

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

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

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

Now, calculate the number of moles of Co(NO₃)₂. 6H₂O:

$\backslash text\{Moles\; of\; solute\}\; =\; \backslash frac\{\backslash text\{Mass\; of\; solute\}\}\{\backslash text\{Molar\; mass\; of\; solute\}\}\; =\; \backslash frac\{30\; \backslash text\{\; g\}\}\{291\; \backslash text\{\; g\; mol\}^\{-1\}\}\; \backslash approx\; 0.1031\; \backslash text\{\; mol\}$

The volume of the solution is given as 4.3 L.

Calculate the molarity:

$$Molarity\; (M)\; =\; \backslash frac\{0.1031\; \backslash text\{\; mol\}\}\{4.3\; \backslash text\{\; L\}\}\; \backslash approx\; 0.02398\; \backslash text\{\; M\}$$

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

Rearranging the formula to solve for M₂:

$$M\_2\; =\; \backslash frac\{M\_1V\_1\}\{V\_2\}$$

Substitute the given values:

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

$$M\_2\; =\; \backslash frac\{15\}\{500\}\; \backslash text\{\; M\}\; =\; 0.03\; \backslash text\{\; M\}$$

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.