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Chemistry
Solutions
62 questions, each with its answer.
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Identify the correct statements: I. The molality of 2.5 g of ethanoic acid (Molar mass: 60 g $mol^{-1}$) in 75 g of benzene solution is 0.556 m. II. The molarity of a solution containing 5 g of NaOH (molar mass: 40 g $mol^{-1}$) in 450 mL of solution is 0.278 M at 298 K. III. Aquatic species are more comfortable in cold water. IV. The solubility of gas increases with decrease in pressure. V. For a binary mixture of A and B, the number of moles of A and B are $n_A$ and $n_B$ respectively. The mole fraction of B will be $x_B = \frac{n_A}{n_A + n_B}$ Choose the CORRECT answer from the options given below:I. $m = \frac{n_{solute}}{w_{solvent}(in\ g)} \times 1000 = \frac{2.5}{60 \times 75} \times 1000 = 0.555$ m (True) II. $M = \frac{n_{solute}}{V_{solution}(in\ mL)} \times 1000 = \frac{5}{40 \times 450} \times 1000 = 0.277$ M (True) III. Due to increased solubility of oxygen with decrease of temperature, aquatic species are more comfortable in cold water (True) IV. Solubility of a gas increases with increase in pressure (False) V. $x_B = \frac{n_B}{n_A + n_B}$ (False) So I, II and III only are correct.
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Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to:$CHCl_3 + CH_3COCH_3$ Due to H-bonding between chloroform ($CHCl_3$) and acetone ($CH_3COCH_3$), the A–B attractions are stronger than A–A and B–B, so the solution shows negative deviation from Raoult's law.
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For an ideal solution formed by mixing liquids P and Q, $\Delta_{mix}H=0$ and $\Delta_{mix}V=0$Reason: No interactions occur between P and QAn ideal solution forms when the intermolecular attractive forces P-P and Q-Q are nearly equal to those between P-Q. Since the interactions remain nearly the same, $\Delta_{mix}H=0$ and $\Delta_{mix}V=0$. Interactions between P and Q do occur, so the Assertion is correct and the Reason is incorrect.
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Consider the following statements about the solutions formed by mixing two liquids. A. An ideal solution thus formed obeys Raoult's law throughout the composition range. B. Mixture of chloroform and acetone shows negative deviation from Raoult's law. C. Mixture of aniline and phenol shows positive deviation from Raoult's law.(A) is correct: an ideal solution obeys Raoult's law over the entire composition range. (B) is correct: chloroform and acetone show negative deviation (H-bonding between them). (C) is incorrect: aniline and phenol also show negative deviation from Raoult's law.
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Match List-I with List-II.
Pick the matching item from Column II for each row.
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?
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?
$P_{\text{ideal}} = \chi_X P_X^\circ + \chi_Y P_Y^\circ = \frac{5}{15}(63) + \frac{10}{15}(78) = 21 + 52 = 73\text{ torr}$. 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:By Henry's law, $p = K_H x$; the solubility of a gas is inversely proportional to its Henry's law constant $K_H$. $K_H$: B ($2 \times 10^{-5}$ kbar) < C (35 kbar) C > A.
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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 $mol^{-1} K^{-1}$)$\Pi = CRT$; comparing with $y = mx$, the slope m = RT $T = \frac{slope}{R} = \frac{25.73}{0.083} = 310$ K $= (310 - 273)^\circ C = 37^\circ C$
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Arrange the following compounds in increasing order of their solubilities in chloroform: NaCl, $CH_3OH$, cyclohexane, $CH_3CN$Chloroform is an organic (non-polar/weakly polar) solvent, so non-polar/covalent compounds are more soluble in it, while ionic NaCl is least soluble. As dipole moment of the solute decreases, solubility in chloroform increases. Order: $NaCl<CH_3CN<CH_3OH<$ Cyclohexane.
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Mass of glucose ($C_6H_{12}O_6$) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L$^{-1}$ solution of urea ($NH_2CONH_2$) is (Given: Molar mass in g mol$^{-1}$ C: 12, H: 1, O: 16, N: 14)For isotonic solutions, osmotic pressures are equal, so molar concentrations are equal: $\dfrac{m_{glucose}}{M_{glucose}}=\dfrac{m_{urea}}{M_{urea}}\Rightarrow\dfrac{m}{180}=\dfrac{15}{60}\Rightarrow m=\dfrac{180}{4}=45$ g.
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Helium is used to dilute oxygen in diving apparatus.Reason: Helium has high solubility in $O_2$
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. -
The density of 1 M solution of a compound 'X' is 1.25 g mL$^{-1}$. The correct option for the molality of solution is (Molar mass of compound X = 85 g):$m=\dfrac{1000\times M}{1000\times d-M\times M_w}=\dfrac{1000\times1}{1000\times1.25-1\times85}=\dfrac{1000}{1165}=0.858$ m.
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Which amongst the following aqueous solution of electrolytes will have minimum elevation in boiling point? Choose the correct option:-Boiling point elevation $\propto i\times M$ (van't Hoff factor times molar concentration). Comparing: NaCl (0.05M): $2\times0.05=0.1$; KCl (0.1M): $2\times0.1=0.2$; $MgSO_4$ (0.1M): $2\times0.1=0.2$; NaCl (1M): $2\times1=2$. The minimum value ($i\times M=0.1$) is for 0.05 M NaCl, giving the minimum elevation in boiling point.
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In one molal solution that contains 0.5 mole of a solute, there is$m = \frac{moles\ of\ solute}{weight\ of\ solvent\ (g)} \times 1000$ $1 = \frac{0.5}{w} \times 1000$ Weight of solvent = 500 g
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$K_H$ value for some gases at the same temperature 'T' are given: Ar: 40.3 kbar; $CO_2$: 1.67 kbar; HCHO: $1.83\times10^{-5}$ kbar; $CH_4$: 0.413 kbar where $K_H$ is Henry's Law constant in water. The order of their solubility in water is:Solubility of a gas $\propto\frac{1}{K_H}$. Higher $K_H$ means lower solubility: Ar (40.3) < $CO_2$ (1.67) < $CH_4$ (0.413) < HCHO ($1.83\times10^{-5}$).
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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:$\pi = iCRT$ with i = 1 for all three (non-electrolytes) and the same volume, so $\pi \propto \frac{mass}{molar\ mass}$ $P_1 = \frac{10}{180}RT$ (glucose), $P_2 = \frac{10}{60}RT$ (urea), $P_3 = \frac{10}{342}RT$ (sucrose) So $P_2 > P_1 > P_3$
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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]$n_B : n_O = 3 : 2 \Rightarrow x_B = \frac{3}{5}$, $x_O = \frac{2}{5}$ $P_s = P_B^\circ x_B + P_O^\circ x_O = 280 \times \frac{3}{5} + 420 \times \frac{2}{5}$ $= 168 + 168 = 336$ mm Hg
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The mixture which shows positive deviation from Raoult's law is:Pure ethanol molecules are hydrogen bonded. On adding acetone, its molecules get in between ethanol molecules and break some of the hydrogen bonds. This weakens the intermolecular attractive interactions, so the mixture shows positive deviation from Raoult's law.
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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):$\Delta T_f = K_f \cdot m = 5.12 \times 0.078$ $= 0.399$ K ≈ 0.40 K
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If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g mol$^{-1}$) of the solute is [Given that molar mass of n-octane is 114 g mol$^{-1}$]$\dfrac{P^0-P_s}{P^0}=\dfrac{n_{solute}}{n_{solute}+n_{solvent}}\approx\dfrac{n_{solute}}{n_{solvent}}$ (dilute solution). Here $\dfrac{P^0-P_s}{P^0}=1-0.8=0.2$. Moles of octane $=\dfrac{114}{114}=1$. $0.2=\dfrac{8/M}{1}\Rightarrow M=\dfrac{8}{0.2}=40\ g\,mol^{-1}$.
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Isotonic solutions have sameIsotonic solutions are defined as solutions having the same osmotic pressure at a given temperature.
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For an ideal solution, the correct option is:For an ideal solution at constant T and P: $\Delta_{mix}H=0$, $\Delta_{mix}V=0$, $\Delta_{mix}S>0$, $\Delta_{mix}G<0$ So the correct option is $\Delta_{mix}H=0$.
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The mixture that forms maximum boiling azeotrope is:Solutions showing negative deviation from Raoult's law form maximum boiling azeotropes. Water and nitric acid show negative deviation, so they form a maximum boiling azeotrope. (Ethanol + water and acetone + carbon disulphide show positive deviation; heptane + octane is nearly ideal.)
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Which of the following statements is correct regarding a solution of two compounds A and B exhibiting positive deviation from ideal behaviour?A solution shows positive deviation from ideal behaviour when the A-A and B-B intermolecular attractions are stronger than the A-B attractions.
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