🧪 Class Test – Chapters: Solutions, Electrochemistry & Chemical Kinetics: Answers

 

✅ Answers

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Q1.
Given:
Mass of NaCl = 5.85 g
Molar mass of NaCl = 58.5 g/mol → moles = 0.1 mol
Mass of water = 100 g → moles of water = 100/18 = 5.56 mol
Vapour pressure of pure water $P^0 = 23.8$ mmHg

NaCl dissociates into Na⁺ and Cl⁻ → Van’t Hoff factor $i = 2$

Relative lowering in vapour pressure =

$$\Delta P = i \cdot \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \cdot P^0 \\ = 2 \cdot \frac{0.1}{0.1 + 5.56} \cdot 23.8 \approx 0.83 \text{ mm Hg}$$

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Q2.
Standard EMF:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \\ = 0.34\ \text{V} - (-0.76\ \text{V}) = 1.10\ \text{V}$$

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Q3.
Rate law: $R = k[A]^2[B]$
Order = 2 (from A) + 1 (from B) = 3

Change:
$[A] \rightarrow 2[A],\ [B] \rightarrow \frac{1}{2}[B]$
New rate = $k(2[A]{)}^2(\frac{1}{2}[B])=4k[A{]}^2\cdot \frac{1}{2}[B]=2k[A{]}^2[B]$

So, rate doubles.

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Q4.
Van’t Hoff factor (i) = Actual number of particles / Expected number of particles
It helps to determine colligative properties when a solute undergoes dissociation or association.
For electrolytes:

$$i = 1 + \alpha(n - 1)$$

(where $\alpha$ = degree of dissociation, n = number of particles)

So, $\mathrm{i<span\; class="katex">is\; used\; to\; calculate\; the</span><strong\; data-end="1522"\; data-start="1496">degree\; of\; dissociation</strong>.}$

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Q5.
Zero-order reaction:
Rate law: $\text{Rate} = k$
Unit of rate = mol L⁻¹ s⁻¹
So, units of $k$ = mol L⁻¹ s⁻¹

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Q6.
Mass of glucose = 18 g
Molar mass = 180 g/mol → moles = 0.1 mol
Mass of water = 100 g → moles = 5.56 mol

Relative lowering in vapour pressure:

$$\frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} = \frac{0.1}{0.1 + 5.56} \approx 0.0177$$

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Q7.

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \\ = 0.34\ \text{V} - (-0.44\ \text{V}) = 0.78\ \text{V}$$

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Q8.
Rate law: $R = k[A][B]^2$
Order = 1 + 2 = 3

Change:

$$[A] \rightarrow 2[A],\ [B] \rightarrow 2[B] \\ R_{\text{new}} = k(2[A])(2[B])^2 = k \cdot 2[A] \cdot 4[B]^2 = 8k[A][B]^2 = 8 \cdot R$$

So, rate increases 8 times.

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Q9.
Abnormal molar mass is observed when a solute dissociates or associates in solution, causing deviation from expected molar mass.

Correct molar mass:

$$\text{Observed molar mass} = \frac{\text{Normal molar mass}}{i}$$

So, Van’t Hoff factor helps adjust for abnormal behaviour and calculate true molar mass.

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Q10.
For first-order reaction:
Rate = $k[A]<span\; class="mord\; mathnormal"><strong\; data-end="2786"\; data-start="2772">units\; of\; k</strong>\; =\; s⁻¹</span>$

Rate law = $\text{Rate} = k[H_2O_2]$