Molality of 0.8 M$0.8\,\mathrm{M}$ H₂SO₄$\mathrm{H}_2\mathrm{SO}_4$ solution (density 1.06 g cm⁻³$1.06\,\mathrm{g\,cm}^{-3}$) is **815** × 10⁻³ m$\times 10^{-3}\,\mathrm{m}$.

Core Logic

A system demonstrating a negative deviation from Raoult's law implies tighter molecular attractions between components (A-B$A-B$ interactions are stronger than A-A$A-A$ or B-B$B-B$). This decreases the aggregate escaping tendency, yielding a decreased total vapour pressure. Consequently, a higher thermal energy threshold is required to reach the boiling threshold, causing an increased boiling point.

Pattern Recognition

Negative deviation arrow$\rightarrow$ Vapour Pressure drops arrow$\rightarrow$ Boiling Point rises inversely.

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Class 12 Chemistry: Solutions

Related Formula

Core Logic

Let's assume we have 100 g of the solution. Mass of H₂SO₄$H_2SO_4$ in 100 g solution = 31.4 g$31.4\text{ g}$. Moles of H₂SO₄$H_2SO_4$ (nsolute$n_{\text{solute}}$) = (31.4)/(98) mol$\frac{31.4}{98}\text{ mol}$.

Volume of the solution (V$V$) can be found using density:

Step 1: Calculating Molarity

Molarity is defined as moles of solute per liter of solution:

Rounding off to the nearest integer gives 4.

Pattern Recognition

Whenever percentage by mass (w/w$w/w$) and density (d$d$ in g/mL) are given, use the direct formula: M = (%(w/w) × d × 10)/(Mw)$M = \frac{\%(w/w) \times d \times 10}{M_w}$. This saves enormous time.

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Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

π = C R T$\pi = C R T$

where π$\pi$ is osmotic pressure, C$C$ is molar concentration, R$R$ is gas constant, and T$T$ is absolute temperature.

Core Logic

For a given dilute solution, the concentration (C$C$) and the gas constant (R$R$) are constant. Therefore, osmotic pressure is directly proportional to the absolute temperature. π ∝ T$\pi \propto T$

Step 1: Calculation

Given values: π₁ = 7 × 10⁵ Pa = 70 × 10⁴ Nm⁻²$\pi_1 = 7 \times 10^5 \text{ Pa} = 70 \times 10^4 \text{ Nm}^{-2}$ T₁ = 273 K$T_1 = 273 \text{ K}$ T₂ = 283 K$T_2 = 283 \text{ K}$

Rearranging for π₂$\pi_2$:

Converting to the requested format (× 10⁴ Nm⁻²$\times 10^4 \text{ Nm}^{-2}$):

Rounding off yields 72.56 (or 73 depending on required decimal places).

Chapter Mix

Class 12 Chemistry: Solutions

Related Formula

Core Logic

The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the solution.

Total number of moles in the solution = nA + nB + nC$= n_A + n_B + n_C$

Mole fraction of C (χC$\chi_C$) = (nC)/(nA + nB + nC)$\frac{n_C}{n_A + n_B + n_C}$

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Class 12 Chemistry: Solutions