Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ₂$PQ_{2}$. When 1 g of PQ is dissolved in 50 g of solvent ‘A’. Δ Tb$\Delta T_{b}$ was 1.176 K while when 1 g of PQ₂$PQ_{2}$ is dissolved in 50 g of solvent ‘A’, Δ Tb$\Delta T_{b}$ was 0.689 K. (Kb$K_{b}$ of ‘A’ = 5 K kg mol⁻¹$5\text{ K kg mol}^{-1}$). The molar masses of elements P and Q (in g mol⁻¹$\text{g mol}^{-1}$) respectively, are :

Related Formula

π = i C R T$\pi = i C R T$

Core Logic

Given:

• π = 12 atm$\pi = 12 \text{ atm}$
• T = 300 K$T = 300 \text{ K}$
• i = 2$i = 2$ (for complete dissociation of NaCl$\text{NaCl}$)
• R = 0.08 L atm K⁻¹ mol⁻¹$R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$

Step 1: Calculating Strength

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol$\text{NaCl} = 23 + 35.5 = 58.5 \text{ g/mol}$. Strength = 0.25 × 58.5 = 14.625 g/L ≈ 15 g/L$= 0.25 \times 58.5 = 14.625 \text{ g/L} \approx 15 \text{ g/L}$.

Pattern Recognition

Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor i = 2$i = 2$ for NaCl$\text{NaCl}$.

Chapter Mix

Class 12 Chemistry: Solutions

Related Formula

Core Logic

Using boiling point elevation:

Moles of solvent Y = (150)/(300) = 0.5 mol$= \frac{150}{300} = 0.5 \text{ mol}$. Moles of solute X = 0.1 × 0.150 = 0.015 mol$= 0.1 \times 0.150 = 0.015 \text{ mol}$.

Relative lowering of vapour pressure = χsolute = (0.015)/(0.015 + 0.5) ≈ (0.015)/(0.5) = 0.03 = 3 × 10⁻²$= \chi_{\text{solute}} = \frac{0.015}{0.015 + 0.5} \approx \frac{0.015}{0.5} = 0.03 = 3 \times 10^{-2}$.

Step 1: Final Calculation

Thus, the nearest integer value is 3.

Pattern Recognition

Sees: boiling point elevation combined with relative lowering of vapour pressure calculation. Trap: Confusing solvent mass with solution mass when calculating mole fraction.

Chapter Mix

Class 12 Chemistry: Solutions

Related Formula

Taking logarithm on both sides:

Core Logic

From Henry's Law:

This equation represents a straight line of the form y = mx + c$y = mx + c$, where: y = P(g)$y = \log P(g)$ x = X(g)$x = \log X(g)$ Slope (m$m$) = +1 y-intercept (c$c$) = KH$\log K_{H}$ (which is positive).

The plot of P$\log P$ versus X$\log X$ is a straight line with a positive slope and a positive intercept on the y-axis.

Step 1: Conclusion

Graph 3 corresponds to a straight line with a positive slope and a positive y-intercept.

Pattern Recognition

When variables are multiplied (P = KH X$P = K_{H} X$), their log plot is always a straight line with a slope of +1 and an intercept equal to the log of the constant.

Chapter Mix

Class 12 Chemistry: Solutions

Core Logic

Statement I: KH$K_H$ is independent of concentration as long as the solution behaves ideally (is very dilute). This matches the physical definition of Henry's Law constant. (True)

Statement II: Henry's Law constant (KH$K_H$) depends on the nature of the gas, the nature of the solvent, and the temperature. Therefore, it absolutely differs for the same gas dissolved in different solvents. (False)

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

The constants for solubility laws (like KH$K_H$) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent.

Chapter Mix

Class 12 Chemistry: Solutions