Solutions

$∆{\mathrm{T}}_{\mathrm{b}} = {\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$

$∆{\mathrm{T}}_{\mathrm{b}} \mathrm{\alpha } \mathrm{i} \times \mathrm{m}$

For NaCl " i " = 2 and Urea " i " = 1

Higher the product of vant Hoff factor and concentration higher the boiling point.

i × C = (i) 2 × 10^–4 (ii) 1 × 10^–4 (iii) 2 × 10^–3 (iv) 2 × 10^–2

Hence the order for boiling point is: (ii) < (i) < (iii) < (iv)

• Density of solution = 1.12 g/mL

• Molality (m) = 3 mol/kg

• Molar mass of NaOH = 40 g/mol

• Molarity (x) = ?

We are to find the molarity $x$ (in mol/L) of NaOH.

Molality $m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}}$

Given:

$$\text{Molality} = 3 \, \mathrm{mol/kg}$$

Let’s assume we have 1 kg (1000 g) of water. Then the number of moles of NaOH = 3 mol.

Mass of NaOH =

$$3 \, \text{mol} \times 40 \, \mathrm{g/mol} = 120 \, \mathrm{g}$$

So total mass of solution =

$$1000 \, \mathrm{g} \text{ (water)} + 120 \, \mathrm{g} \text{ (NaOH)} = 1120 \, \mathrm{g}$$

Given density = 1.12 g/mL

So for 1120 g of solution:

$$\text{Volume} = \frac{1120 \, \mathrm{g}}{1.12 \, \mathrm{g/mL}} = 1000 \, \mathrm{mL} = 1 \, \mathrm{L}$$

Molarity $x = \frac{\text{mol of solute}}{\text{volume of solution in L}}$

We have 3.0 mol NaOH in 1 L of solution

x = 3.0

Mixing polar acetone with non-polar CS₂ weakens intermolecular attractions (A–B < A–A/B–B) → positive deviation from Raoult’s law.
The other pairs (acetone–chloroform, acetone–aniline, chloroform–benzene) form specific associations/H-bonding → negative deviation.

Calculate Van't Hoff Factor (i) for $A{X}_2$

$$\begin{gathered} \Delta T_b=i\times K_b\times m \\ 0.0156=i\times 0.52\times 0.01 \\ i=\frac{0.0156}{0.0052}=3 \end{gathered}$$

Since $i=3,A{X}_2$ fully ionises
Calculate Van't Hoff Factor (i) for $A{Y}_2$

$$\begin{gathered} \Delta T_b=i\times K_b\times m \\ 0.0260=i\times 0.52\times 0.0508 \\ i=\frac{0.0260}{0.0264}=\approx 1 \end{gathered}$$

Since $i\approx 1,A{Y}_2$ does not ionise (remains unionised).

According to Henry’s law,
Lower K_H means higher solubility.
Given:

• Kₕ(B) = 2 × 10⁻⁵ (lowest) → highest solubility

• Kₕ(C) = 35

• Kₕ(A) = 145 (highest) → lowest solubility
  So, B > C > A.

$\mathrm{Molarity} \mathrm{of} \mathrm{Mixture} =\frac{{\mathrm{M}}_{1 }{\mathrm{V}}_1+ {\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}= \frac{200\times 0.2 + 400 \times 0.5}{600}=0.4\mathrm{M}$

- Mass of $\mathrm{NaCl}=5.85 \mathrm{g}$
- Volume of water $$$=500 \mathrm{mL}=0.5 \mathrm{L}$
- Molar Mass of $\mathrm{Na}\left(\mathrm{Na}\right)=23 \mathrm{g}/\mathrm{mol}$
- Molar Mass of $\mathrm{Cl}\left(\mathrm{Cl}\right)=35.5 \mathrm{g}/\mathrm{mol}$
1. Calculate the molar mass of NaCl :
- Molar mass of $\mathrm{NaCl}=23+35.5=58.5 \mathrm{g}/\mathrm{mol}$
2. Calculate the number of moles of $\mathbf{NaCl}$ :
- Number of moles = mass of$\mathrm{NaCl}/$ molar mass of NaCl \(=5.85 \mathrm{~g} / 58.5 \mathrm{~g} / \mathrm{mol}\) \(=0.1\) moles
3. Convert the volume of the solution from mL to L :
- Volume \(=\mathbf{5 0 0 ~ m L}=\mathbf{0 . 5 ~ L}\)
4. Calculate the molarity (M):
- Molarity $\left(\mathrm{M}\right)=$ number of moles of solute \(/\) volume of solution in liters \(=0.1\) $=0.2\mathrm{M}$

Given 30 g HNO₃, 75% w/w solution, and density = 1.25 g/mL, total solution mass is:

$$\frac{30\times 100}{75}=40\text{ g}$$

Using Density = Mass/Volume,

$$V=\frac{40}{1.25}=32\text{ mL}$$
Final answer: 32 mL.

Using Raoult’s Law,

$$\frac{\mathrm{\Delta }P}{P_0}=\text{mole fraction of solute}(X_B)$$

Given, ΔP = 10 mm Hg and X_B = 0.2, so P₀ = 10/0.2 = 50 mm Hg.

For ΔP = 20 mm Hg,

$$X_B=\frac{20}{50}=0.4$$

Thus, mole fraction of solvent (X_A) = 1 - 0.4 = 0.6.

$$\begin{gathered} {\mathrm{A}}_2\mathrm{B} \to 2{\mathrm{A}}^{\mathrm{x}+}+{\mathrm{B}}^{\mathrm{y}-} \\ \mathrm{Degree} \mathrm{of} \mathrm{dissociation}\left(\mathrm{\alpha }\right) =30\% =0.3, \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{ions} \left(\mathrm{n}\right) =3 \\ \mathrm{Now} \mathrm{for} \mathrm{dissociation} \mathrm{\alpha }=\frac{\mathrm{i}-1}{\mathrm{n}-1} (\mathrm{i} = \mathrm{van}\text{'}\mathrm{t} \mathrm{Hoff} \mathrm{factor}) \\ 0.3=\frac{\mathrm{i}-1}{3-1}\Rightarrow \mathrm{i}-1 =0.6 , \mathrm{Hence} \mathrm{i}= 1.60 \end{gathered}$$

By the Rault's law

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}} ={\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}} .... \left(1\right) \\ {\mathrm{P}}_{\mathrm{B}} ={\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{B}} .....\left(2\right) \\ \mathrm{Now} \\ {\mathrm{P}}_{\mathrm{A}} ={\mathrm{Y}}_{\mathrm{A}} {\mathrm{P}}_{\mathrm{T}} \Rightarrow {\mathrm{P}}_{\mathrm{T}} = \frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}} \\ {\mathrm{P}}_{\mathrm{B}} ={\mathrm{Y}}_{\mathrm{B}} {\mathrm{P}}_{\mathrm{T}} \Rightarrow {\mathrm{P}}_{\mathrm{T}} = \frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{B}}} \\ \frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}} =\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{B}}} \\ \frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}} =\frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{B}}} \left(\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\right) \\ \frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}} =\frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}\left(1-{\mathrm{X}}_{\mathrm{A}}\right)}{1-{\mathrm{Y}}_{\mathrm{A}}} \\ \frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}\left(1-{\mathrm{Y}}_{\mathrm{A}}\right)}{{\mathrm{Y}}_{\mathrm{A}}} =\frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}\left(1-{\mathrm{X}}_{\mathrm{A}}\right)}{{\mathrm{X}}_{\mathrm{A}}} \\ \frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}}{{\mathrm{Y}}_{\mathrm{A}}} -\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}} {\mathrm{Y}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}}=\frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{A}}}-\frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}} {\mathrm{X}}_{\mathrm{A}}}{{\mathrm{X}}_{\mathrm{A}}} \\ \frac{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{A}}}=\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}}{{\mathrm{Y}}_{\mathrm{A}}}+\left({\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}\right) \\ \frac{1}{{\mathrm{X}}_{\mathrm{A}}}=\frac{1}{{\mathrm{Y}}_{\mathrm{A}}}\left(\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}}{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}}\right)+\frac{\left({\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}\right)}{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}} \\ \mathrm{Compare} \mathrm{with} \mathrm{y} = \mathrm{mx} + \mathrm{c} \end{gathered}$$

Mixing polar acetone with non-polar CS₂ weakens intermolecular attractions (A–B < A–A/B–B) → positive deviation from Raoult’s law.
The other pairs (acetone–chloroform, acetone–aniline, chloroform–benzene) form specific associations/H-bonding → negative deviation.

Boiling point elevation depends on molality × van’t Hoff factor (i). NaCl (i = 2) dissociates, while glucose (i = 1) does not. Thus, NaCl solution shows higher elevation.

$$\begin{gathered} molality \left(m\right) =\frac{1000M}{1000d -M\times M_0} (where M= molarity , M_{0 }=molar mass) \\ m =\frac{3000}{1250 -58.5\times 3} =2.79 \end{gathered}$$

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}} = \mathrm{i} \times {\mathrm{K}}_{\mathrm{b}}\times \mathrm{m} \\ \mathrm{Higher} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{i}\times \mathrm{m}, \mathrm{higher} \mathrm{will} \mathrm{the} \mathrm{BP} \\ \mathrm{for} \mathrm{NaCl} . \mathrm{i}= 2 \mathrm{and} \mathrm{for} \mathrm{urea} \mathrm{i}= 1 \\ \mathrm{Hence} , \mathrm{the} \mathrm{order} \mathrm{of} \mathrm{BP} \mathrm{is} \mathrm{iii}>\mathrm{ii}>\mathrm{i}>\mathrm{iv} \end{gathered}$$

Show no change in volume since solution is 0.9%

$$\begin{gathered} molality \left(m\right) =\frac{1000M}{1000d -M\times M_0} (where M= molarity , M_{0 }=molar mass) \\ m =\frac{3000}{1250 -58.5\times 3} =2.79 \end{gathered}$$

$$\begin{gathered} \mathrm{Depression} \mathrm{in} \mathrm{freezing} \mathrm{point} ∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{f}}.\mathrm{m} \\ \mathrm{molality} \mathrm{m}=\frac{1}{256}\times \frac{1000}{50}=0.0781 \\ {\mathrm{K}}_{\mathrm{f}}=\frac{∆{\mathrm{T}}_{\mathrm{b}}}{\mathrm{m}}=0.4/0.0781=5.12 \mathrm{KKg} /\mathrm{mol} \end{gathered}$$

Molality \((m)\) is given as

$m=\frac{n_{\text{solute }}}{k{g}_{\text{solvent }}}$

To calculate no. of moles of solute we take the help of molarity and to determine mass of solvent we take the help of density of solution. \(3 M\) solution of \(N a C l\) implies that there are 3 moles of \(N a C l\) in every 1 litre of solution. Considering \(1 L\) of solution, we have

$n_{\mathrm{NaCl}}=3 \mathrm{mol}$

Mass of \(1 L\) solution of \(\mathrm{NaCl}=(\) volume \()(\) density )

$\begin{array}{rl}& =\left(1000\mathrm{mL}\right)\left(1.25{\mathrm{gmol}}^{-1}\right)\\ & =1250\mathrm{g}\end{array}$

Mass of NaCl in 1L of solution $\begin{array}{rl}& =\left(n_{\mathrm{NaCl}}\right)\left({\mathrm{mm}}_{\mathrm{NaCl}}\right)\end{array}$

$\begin{array}{rl}& =\left(3\mathrm{mol}\right)\left(58.5{\mathrm{gmol}}^{-1}\right)\\ & =175.5\mathrm{g}\end{array}$

Mass of solvent $\left({\mathrm{H}}_2\mathrm{O}\right)=$ Mass of solution - Mass of solute $=(1250 \mathrm{g})-(175.5 \mathrm{g})$

$=1074.5g$

Thus

$\begin{array}{rl}& m=\frac{n_{\text{solute}}}{g_{\text{solv}}}\frac{1000g}{kg}\\ & =\frac{3\mathrm{mol}}{1074.5g}\times \frac{1000g}{\mathrm{kg}}\\ & 2.79 mol k{g}^{-1}\end{array}$

Alternatively $m=\frac{M_{\text{soln }}}{1000d-M_{\text{soln }m{m}_{\text{solute }}}}\times 1000$

$\begin{array}{rl}& =\frac{3}{\left(1000\right)\left(1.25\right)-\left(3\right)\left(58.5\right)}\times 1000\\ & =2.79{\mathrm{molkg}}^{-1}\end{array}$

$\begin{array}{l}\mathrm{M}=\frac{{\mathrm{w}}_2\times 1000}{{\mathrm{M}}_2\times \mathrm{V}}\\ \frac{1}{20}=\frac{{\mathrm{w}}_2\times 1000}{180\times 250}\end{array}$

${\mathrm{w}}_2=\frac{180\times 250}{20\times 1000}=2.25\mathrm{g}$

$\pi =CRT$

Slope = RT

25.73 = RT

$\mathrm{T}=\frac{25.73}{0.083}=310\mathrm{K}$

$\therefore$ Temperature in ºC = 310 – 273

= 37ºC

When NaCl (sodium chloride, or table salt) is added to ice at $0°C$ in an ice cream box, it lowers the freezing point of the water, effectively preventing the ice cream from melting by creating a colder mixture around the container; this phenomenon is called "freezing point depression"

Step 1: Calculate the moles of solute Given:
- Mass of solute = 1g
- Molar mass of solute = 256 g/mol

Moles of solute $=\frac{1}{256}=0.00390625$ moles.
Step 2: Calculate the molality
Mass of solvent $=50g=0.050kg$

$m=\frac{0.00390625}{0.050}=0.078125mol/kg$

Step 3: Calculate $K_f$
Using the formula:

$$\begin{gathered} K_f=\frac{\Delta T_f}{m}=\frac{0.40}{0.078125} \\ {K}_f\approx 5.12Kkgmo{l}^{-1} \end{gathered}$$

Molality-Moles of solute/mass of solvent in kg ∵ Molality does not involve volume term. ∴ It is independent of temperature.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}} = \mathrm{i} \times {\mathrm{K}}_{\mathrm{b}}\times \mathrm{m} \\ \mathrm{Higher} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{i}\times \mathrm{m}, \mathrm{higher} \mathrm{will} \mathrm{the} \mathrm{BP} \\ \mathrm{for} \mathrm{NaCl} . \mathrm{i}= 2 \mathrm{and} \mathrm{for} \mathrm{urea} \mathrm{i}= 1 \\ \mathrm{Hence} , \mathrm{the} \mathrm{order} \mathrm{of} \mathrm{BP} \mathrm{is} \mathrm{iii}>\mathrm{ii}>\mathrm{i}>\mathrm{iv} \end{gathered}$$

$\mathrm{Molarity} \mathrm{of} \mathrm{Mixture} =\frac{{\mathrm{M}}_{1 }{\mathrm{V}}_1+ {\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}= \frac{200\times 0.2 + 400 \times 0.5}{600}=0.4\mathrm{M}$

If an endothermic reaction is non-spontaneous at the freezing point and becomes feasible at its boiling point, then: ΔH is +ve and ΔS is +ve.