At T(K)$T(\text{K})$, 100 g$100\text{ g}$ of 98% H₂SO₄ (w/w)$98\% \text{ H}_2\text{SO}_4\text{ (w/w)}$ aqueous solution is mixed with 100 g$100\text{ g}$ of 49% H₂SO₄ (w/w)$49\% \text{ H}_2\text{SO}_4\text{ (w/w)}$ aqueous solution. What is the mole fraction of H₂SO₄$\text{H}_2\text{SO}_4$ in the resultant solution? (Given: Atomic mass H = 1 u$\text{H} = 1\text{ u}$; S = 32 u$\text{S} = 32\text{ u}$; O = 16 u$\text{O} = 16\text{ u}$) (Assume that temperature after mixing remains constant)

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Core Logic

A mixture of carbon disulfide (CS₂$CS_2$) and acetone (CH₃-CO-CH₃$CH_3-CO-CH_3$) shows a positive deviation from Raoult's Law. This happens because the dipole-dipole interactions between acetone and CS₂$CS_2$ are weaker than the attractive interactions between the pure molecules themselves.

Step 1: Graph Interpretation

Due to positive deviation, the actual partial pressure curve will bulge upwards (convex) relative to the ideal straight line (Raoult's Law). The first graph precisely depicts this upward curve starting from zero and reaching P^°CS₂$P^\circ_{CS_2}$.

Pattern Recognition

Acetone + Carbon disulfide = Positive Deviation. Positive deviation means vapor pressure is higher than expected, so the curve sags upwards above the straight-line prediction.

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Core Logic

Case 1: Moles of A (nA$n_A$) = 3, Moles of B (nB$n_B$) = 1. Total moles = 3 + 1 = 4$3 + 1 = 4$. Mole fraction of A (XA$X_A$) = (3)/(4)$\frac{3}{4}$, Mole fraction of B (XB$X_B$) = (1)/(4)$\frac{1}{4}$. Vapour pressure (Pₛ$P_{\text{s}}$) = 500 mm Hg$500 \text{ mm Hg}$. Using Raoult's law:

Step 1: Setting up Case 2

Case 2: 1 mol of A is further added. So, new moles of A (nA$n_A^{\prime}$) = 4. Moles of B remains 1. Total moles = 4 + 1 = 5$4 + 1 = 5$. Mole fraction of A (XA$X_A^{\prime}$) = (4)/(5)$\frac{4}{5}$, Mole fraction of B (XB$X_B^{\prime}$) = (1)/(5)$\frac{1}{5}$. Vapour pressure increases by 20 mm Hg$20 \text{ mm Hg}$, so new Pₛ = 500 + 20 = 520 mm Hg$P_{\text{s}} = 500 + 20 = 520 \text{ mm Hg}$. Using Raoult's law:

Step 2: Solving for Pure Vapour Pressures

Subtract Equation 1 from Equation 2:

Substitute PA°$P_A^{\circ}$ back into Equation 1:

Pattern Recognition

A standard two-equation Raoult's law system. Cross-multiply out the denominators immediately to get simple linear equations (Ax + By = C$Ax + By = C$) which allow straightforward elimination.

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where, Ctotal = nA + nBVsolution (L)$C_{\text{total}} = \frac{n_A + n_B}{V_{\text{solution (L)}}}$

Core Logic

Mass of solute 'A' = 0.3 g$0.3 \text{ g}$ Moles of solute 'A' = 0.3 g60 g/mol = (1)/(200) mol$\frac{0.3 \text{ g}}{60 \text{ g/mol}} = \frac{1}{200} \text{ mol}$

Mass of solute 'B' = 0.9 g$0.9 \text{ g}$ Moles of solute 'B' = 0.9 gm180 g/mol = (1)/(200) mol$\frac{0.9 \text{ gm}}{180 \text{ g/mol}} = \frac{1}{200} \text{ mol}$

Total moles = (1)/(200) + (1)/(200) = (2)/(200) mol$\frac{1}{200} + \frac{1}{200} = \frac{2}{200} \text{ mol}$

Volume of solution = 100 mL = 0.1 L$100 \text{ mL} = 0.1 \text{ L}$

Total molarity (C$C$) of all solutes:

Step 1: Calculate Osmotic Pressure

π = C RT$\pi = C RT$

Pattern Recognition

For non-electrolytes, i=1$i=1$. Add the moles of all solutes present to find the total effective molarity. Use T = 300 K$T = 300 \text{ K}$ for 27°C$27^{\circ}\text{C}$.

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Core Logic

From Relative Lowering of Vapour Pressure: P° = 640 mm Hg$P^{\circ} = 640 \text{ mm Hg}$ Pₛ = 600 mm Hg$P_s = 600 \text{ mm Hg}$ Δ P = 40 mm Hg$\Delta P = 40 \text{ mm Hg}$

Moles of solute n = (W)/(M)$n = \frac{W}{M}$ Mole fraction Xsolute = Δ PP° · (1)/(i)$X_{\text{solute}} = \frac{\Delta P}{P^{\circ}} \cdot \frac{1}{i}$ (for dilute solutions, or properly i · Xsolute = Δ PP°$i \cdot X_{\text{solute}} = \frac{\Delta P}{P^{\circ}}$ as given in the pdf approach) Δ PP° = i · Xsolute Xsolute = (40)/(640) × (1)/(i)$\frac{\Delta P}{P^{\circ}} = i \cdot X_{\text{solute}} \implies X_{\text{solute}} = \frac{40}{640} \times \frac{1}{i}$

Now, from Boiling Point Elevation: Δ Tb = 375 - 373 = 2 K$\Delta T_b = 375 - 373 = 2 \text{ K}$ m = moles of solutemass of solvent in kg = (W/M)/(100/1000) = (W)/(M) × 10$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W/M}{100/1000} = \frac{W}{M} \times 10$ Δ Tb = i × Kb × m$\Delta T_b = i \times K_b \times m$ 2 = i × 0.52 × ( (W/M)/(100) × 1000 )$2 = i \times 0.52 \times \left( \frac{W/M}{100} \times 1000 \right)$ i = (2)/(5.2) × (M)/(W)$i = \frac{2}{5.2} \times \frac{M}{W}$

Step 1: Calculate Mole Fraction

Substitute the value of i$i$ into the Xsolute$X_{\text{solute}}$ equation:

Pattern Recognition

When an unknown electrolyte is involved, isolate the van't Hoff factor (i$i$) from one colligative property equation and substitute it into the other to eliminate i$i$ and solve for the target variable.

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