SolutionsNEET Chemistry · Class 12 · NCERT Chapter 1

Solution

The freezing point depression depends on the number of particles in solution. ${\text{Al}}_2({\text{SO}}_4{)}_3$ dissociates into 5 ions per formula unit, leading to the largest number of particles and thus the largest freezing point depression, so option (c) is correct.

Solution

1.0 m solution means 1 mole solute is present in 1000 g water. nH2O=55.5 mol H2O XSolute= nSol ute nSolute+nH2O = 1 1+ 55.5=0.0177

Solution

16.9 g AgNO3 is present in 100 mL solution. ∴8.45 g AgNo3 is present in 50 mL solution 5.8 g Na Cl is present in 100 mL solution ∴2.9 g NaCl is present in 50 mL solution Mass of AgCl precipitated =0.049×143.5 g =7 g A gCl

Solution

A – benzene, B – Toluene 𝑃𝑇 = 𝑃𝐴 𝑜 𝑋𝐴 + 𝑃𝐵 𝑜 𝑋𝐵 = 12.8 × 0.5 + 3.85× 0.5 = 6. 2 + 1.925 = 8.125 Also, mole fraction of benzene in vapour form 𝑌𝐴 = 𝑃𝐴 𝑜 𝑋𝐴 𝑃𝑇 = 6.2 8.121 = 0.75 And mole fraction of Toluene in vapour form 𝑌𝐵 = 1 − 0.75 = 0.25

Solution

At 100𝑜𝐶 (boiling point) Vapour pressure of water 𝑃𝑜 = 𝑃𝑎𝑡𝑚 = 760 𝑚𝑙 www.vedantu.com 44 ∴ 𝑃𝑜−Ps Po = Xsolute ⇒ 760−732 760 = nsolute nsolvent ⇒ 28 760 = 6.5 m⁄ 100 18⁄ ⇒ m = 6.5×18×760 28×100 ≈ 32 Now, ∆Tb = Kb molality = 0.52× 6.5 32⁄ 0.1 = 0.52×6.5 32×0.1 = 1.05 ≈ 1s ∴ Boiling point of solution = 100+ 1 =101oC

Solution

Kf (molal depression constant) is a characteristic of solvent and is independent of molality.

Solution

Molarity includes volume of solution which can change with change in temperature.

Solution

To find the solution with pH = 1, calculate the concentration of ${\text{H}}^{+}$ ions. For option (a), $[{\text{H}}^{+}]=\frac{(60\times 10)-(40\times 10)}{100}=2\text{M}$. The pH is $-\log [{\text{H}}^{+}]=-\log (2)\approx 0.3$, which is close to 1. For the other options, the ${\text{H}}^{+}$ concentration is either too high or too low. Thus, option (a) is correct.

Solution

Calculate the number of molecules in each case using $n=\frac{m}{M}$ and $N=n\cdot N_A$.
- (a) $n=\frac{0.00224\cdot 1}{0.0821\cdot 273}\approx 0.0001\text{mol}\Rightarrow N\approx 6.022\times 10^{20}$ molecules.
- (b) $n=\frac{0.18}{18}=0.01\text{mol}\Rightarrow N\approx 6.022\times 10^{21}$ molecules.
- (c) $n=\frac{18}{18}=1\text{mol}\Rightarrow N\approx 6.022\times 10^{23}$ molecules.
- (d) $n=10^{-3}\text{mol}\Rightarrow N\approx 6.022\times 10^{20}$ molecules.
Option (c) has the maximum number of molecules, so it is correct.

Solution

Mixture of water and strong acid shows negative deviation.

Solution

Isothermal, Constant external pressure of 2 bar W = – 𝑝𝑒𝑥𝑡.Δ𝑉=−2×(0.25−0.1) bar-litre = –30 joule

Solution

Lindlar’s catalyst converts alkyne into Cis-alkene.

Solution

Ethanol and acetone show positive deviation from Raoult’s law due to weaker intermolecular interactions in the solution compared to pure components. NCERT XII chapter Solutions explains that this results in increased vapor pressure and positive deviation, so option (a) is correct.

Solution

The solubility of ${\text{Ni(OH)}}_2$ in 0.1 M $\text{NaOH}$ can be calculated using the solubility product $K_{sp}=2\times 10^{-15}$. Let $s$ be the solubility of ${\text{Ni(OH)}}_2$. The concentration of ${\text{OH}}^{-}$ from $\text{NaOH}$ is 0.1 M, and from ${\text{Ni(OH)}}_2$ is $2s$. Thus, $K_{sp}=s\cdot (0.1+2s{)}^2\approx s\cdot (0.1{)}^2=2\times 10^{-15}\Rightarrow s=2\times 10^{-13}\text{M}$. Therefore, option (a) is correct.

Solution

The freezing point depression $\Delta T_f$ is given by $\Delta T_f=K_f\cdot m$. Substituting the values, $\Delta T_f=5.12\cdot 0.078=0.40\text{ K}$, so option (c) is correct.

Solution

Osmotic pressure is directly proportional to the molar concentration of solute particles. Glucose and sucrose are non-electrolytes, while urea is a weak electrolyte. The molar masses are ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6=180\text{g/mol}$, ${\text{CH}}_4{\text{N}}_2\text{O}=60\text{g/mol}$, and ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=342\text{g/mol}$. Thus, the order of osmotic pressure is ${\text{P}}_2>{\text{P}}_1>{\text{P}}_3$, so option (a) is correct.

Solution

For an ideal solution, the total vapour pressure is given by $P_{\text{total}}=x_{\text{benzene}}\cdot P_{\text{benzene}}^0+x_{\text{octane}}\cdot P_{\text{octane}}^0$. Substituting the values, $x_{\text{benzene}}=\frac{3}{5}$, $x_{\text{octane}}=\frac{2}{5}$, $P_{\text{benzene}}^0=280\text{mm Hg}$, and $P_{\text{octane}}^0=420\text{mm Hg}$, we get $P_{\text{total}}=\left(\frac{3}{5}\times 280\right)+\left(\frac{2}{5}\times 420\right)=168\text{mm Hg}$. Therefore, option (b) is correct.

Solution

Molality is the moles of solute dissolved per kg of solvent therefore 500 g, 1 molal solution contains 0.5 of solute, as Moles of solutem Mass of solvent (in kg)= 0.51 Mass of solvent (in kg)= ∴ Mass of solvent (in kg) = 0.5 = 500 g

Solution

• Humidity is a solution of liquid in gas
• Alloy is a solution of solid in solid
• Amalgam is a solution of liquid in solid
• Smoke is a solution of solid in gas

Solution

ΔT_b = iK_b × m
ΔT_b ∝ i × m

By considering molarity same as molality

(1) 0.01M Urea i × m = 1 × 0.01 = 0.01
(2) 0.01 M KNO₃ i × m = 2 × 0.01 = 0.02
(3) 0.01M Na₂SO₄ i × m = 3 × 0.01 = 0.03
(4) 0.015M C₆H₁₂O₆ i × m = 1 × 0.015 = 0.015

T_b' = T_b* + ΔT_b

Higher the value of (i × m) more will be the boiling point.

Solution

$P_{total}=X_X\cdot P_X^{\circ }+X_Y\cdot P_Y^{\circ }$

$=\frac{5}{15}\times 63+\frac{10}{15}\times 78$

$=21+52=73\text{ torr}$

Observed total pressure of solution = 70 torr.

Since observed pressure (70 torr) is less than the calculated value (73 torr), the solution shows negative deviation from Raoult's law.