Looking for classes? Ksquare Career Institute, Bengaluru →
Solutions
-
Match List-I with List-II.
Column I
- A. Humidity
- B. Alloys
- C. Amalgams
- D. Smoke
Column II
- I. Solid in solid
- II. Liquid in gas
- III. Solid in gas
- IV. Liquid in solid
Correct answer: A → II, B → I, C → IV, D → III
Humidity: Liquid in gas; Alloys: Solid in solid; Amalgams: Liquid in solid; Smoke: Solid in gas.
Classification of solution types: 1. Humidity represents water vapour dispersed in air (liquid in gas, II). 2. Alloys are homogeneous solid solutions of metals (solid in solid, I). 3. Amalgams are liquid mercury dissolved in a solid metal (liquid in solid, IV). 4. Smoke consists of solid carbon/dust particles dispersed in air (solid in gas, III). Therefore, the correct matching is A-II, B-I, C-IV, D-III. -
Which of the following aqueous solution will exhibit highest boiling point?A $0.01\text{ M Urea}$B $0.01\text{ M KNO}_3$C $0.01\text{ M Na}_2\text{SO}_4$D $0.015\text{ M C}_6\text{H}_{12}\text{O}_6$
Elevation in boiling point $\Delta T_b \propto i \times M$. For $\text{Na}_2\text{SO}_4$, $i \times M = 3 \times 0.01 = 0.03\text{ M}$, which is highest.
Boiling point elevation is a colligative property: $\Delta T_b = i \cdot K_b \cdot m \approx i \cdot K_b \cdot M$. Higher $i \times M$ results in a higher boiling point. Evaluating $i \times M$: (1) Urea: non-electrolyte, $i = 1 \implies 1 \times 0.01 = 0.01$. (2) $\text{KNO}_3$: dissociates into $\text{K}^+ + \text{NO}_3^-$ ($i = 2$) $\implies 2 \times 0.01 = 0.02$. (3) $\text{Na}_2\text{SO}_4$: dissociates into $2\text{Na}^+ + \text{SO}_4^{2-}$ ($i = 3$) $\implies 3 \times 0.01 = 0.03$. (4) Glucose: non-electrolyte, $i = 1 \implies 1 \times 0.015 = 0.015$. Since $0.01\text{ M Na}_2\text{SO}_4$ gives the largest total particle concentration ($0.03\text{ M}$), it exhibits the highest boiling point. -
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of $70\text{ torr}$. The vapour pressures of pure X and Y are $63\text{ torr}$ and $78\text{ torr}$ respectively. Which of the following is true regarding the described solution?A The solution shows positive deviation.B The solution shows negative deviation.C The solution is ideal.D The solution has volume greater than the sum of individual volumes.
$P_{\text{ideal}} = \chi_X P_X^\circ + \chi_Y P_Y^\circ = \frac{5}{15}(63) + \frac{10}{15}(78) = 21 + 52 = 73\text{ torr}$. Since $P_{\text{obs}} = 70\text{ torr} < P_{\text{ideal}}$, the solution shows negative deviation from Raoult's law.
Mole fractions of components: $\chi_X = \frac{5}{5 + 10} = \frac{1}{3}$, $\chi_Y = \frac{10}{15} = \frac{2}{3}$. According to Raoult's law for an ideal solution: $P_{\text{ideal}} = \chi_X P_X^\circ + \chi_Y P_Y^\circ = \left(\frac{1}{3} \times 63\right) + \left(\frac{2}{3} \times 78\right) = 21 + 52 = 73\text{ torr}$. The observed total vapour pressure is $70\text{ torr}$, which is lower than the ideal value ($70\text{ torr} < 73\text{ torr}$). This indicates that intermolecular attractions between X and Y molecules are stronger than X-X and Y-Y interactions, resulting in a negative deviation from Raoult's law. -
The Henry's law constant ($K_H$) values of three gases (A, B, C) in water are 145, $2 \times 10^{-5}$ and 35 kbar, respectively. The solubility of these gases in water follow the order:A A > C > BB A > B > CC B > A > CD B > C > A
-
The plot of osmotic pressure ($\Pi$) vs concentration (mol L$^{-1}$) for a solution gives a straight line with slope 25.73 L bar $mol^{-1}$. The temperature at which the osmotic pressure measurement is done is: (Use $R = 0.083$ L bar $mol^{-1} K^{-1}$)A $25.73^\circ C$B $12.05^\circ C$C $37^\circ C$D $310^\circ C$
-
Helium is used to dilute oxygen in diving apparatus.Reason: Helium has high solubility in $O_2$A Both A and R are true and R is the correct explanation of AB Both A and R are true but R is NOT the correct explanation of AC A is true but R is falseD A is false but R is true
Helium is used because of its very low solubility in blood.
Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood. -
In one molal solution that contains 0.5 mole of a solute, there isA 1000 g of solventB 500 mL of solventC 500 g of solventD 100 mL of solvent
-
The following solutions were prepared by dissolving 10 g of glucose ($C_6H_{12}O_6$) in 250 ml of water ($P_1$), 10 g of urea ($CH_4N_2O$) in 250 ml of water ($P_2$) and 10 g of sucrose ($C_{12}H_{22}O_{11}$) in 250 ml of water ($P_3$). The right option for the decreasing order of osmotic pressure of these solutions is:A $P_1>P_2>P_3$B $P_2>P_3>P_1$C $P_3>P_1>P_2$D $P_2>P_1>P_3$
-
The correct option for the value of vapour pressure of a solution at $45^\circ C$ with benzene to octane in molar ratio 3 : 2 is: [At $45^\circ C$ vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]A 168 mm of HgB 336 mm of HgC 350 mm of HgD 160 mm of Hg
-
The mixture which shows positive deviation from Raoult's law is:A Benzene + TolueneB Acetone + ChloroformC Chloroethane + BromoethaneD Ethanol + Acetone
-
The freezing point depression constant ($K_f$) of benzene is $5.12\ K\,kg\,mol^{-1}$. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):A 0.80 KB 0.40 KC 0.60 KD 0.20 K
-
For an ideal solution, the correct option is:A $\Delta_{mix}S=0$ at constant $T$ and $P$B $\Delta_{mix}V\neq0$ at constant $T$ and $P$C $\Delta_{mix}H=0$ at constant $T$ and $P$D $\Delta_{mix}G=0$ at constant $T$ and $P$
-
The mixture that forms maximum boiling azeotrope is:A Water + Nitric acidB Ethanol + WaterC Acetone + Carbon disulphideD Heptane + Octane