InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
Define the mole fractin of a substance in solution. |
| Answer» For answer, consult Section 2 (Eexpression Concentration of solution) | |
| 852. |
Collingative properties depend onA. the nature of the solute particles dissolved in solution.B. the number of solute particles in solution.C. the physical properties of the solute particles.D. the natre of solvent particles. |
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Answer» Correct Answer - b is the correct answer. |
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| 853. |
Which of the following aqueous solutions should have the highest boiling point?(i) 1.0MNaOH(ii) 1.0MNa2SO4(iii) 1.0MNH4NO3(iv) 1.0MKNO3 |
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Answer» (ii) 1.0MNa2SO4 |
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| 854. |
If two liquids A and B form minimum boiling azeotrope at some specific composition then _______________. (i) A–B interactions are stronger than those between A–A or B–B. (ii) vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution. (iii) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution. (iv) A–B interactions are weaker than those between A–A or B–B. |
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Answer» (i) A–B interactions are stronger than those between A–A or B–B. |
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| 855. |
Which of the following aqueous solutions should have the highest boiling point ?A. `1.0 M NaOH`B. `1.0 M Na_(2)SO_(4)`C. `1.0 M NH_(4)NO_(3)`D. `0.1 M KNO_(3)` |
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Answer» Correct Answer - b Elevation is boiling point temperature `(DeltaT_(b))` is maximym for 1.0 M `Na_(2)SO_(4)` solution since it makes available maximum ins in solution. |
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| 856. |
Which of the following ionic compounds has negative enthalpy of solution?A. `NH_(4)NO_(3)`B. `NH_(4)Cl`C. `KCl`D. `CaCl_(2)` |
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Answer» Correct Answer - 4 Athletes often use instant cold packs and hot packs as first-aid devices to treat injuries. These devices operate by utilizing the enthalpy of solution concept. A typical pack consists of a plastic bag containing a pounch of water and a dry chemical. Stricking the pack causes the pounch to break and the temperature of the pack will be either raised or lowered, depending on whether the enthalpy of solution of the chemical is exothermic or endothermic Generally, calcium chloride or magnesium sulphate is used in hot packs, and ammonium nitrate in cold packs. The reactions are `CaCl_(2)(s)overset(H_(2)O)(rarr) Ca^(2+)(aq.)+2Cl^(-)(aq.), DeltaH_("soln")= -82.8 kJ` `NH_(4)NO_(3)(s)overset(H_(2)O)(rarr) NH_(4)^(+)(aq.)+NO_(3)^(-)(aq.),DeltaH_("soln")= 26.2 kJ` Experiments show that the addition of `40g` of `CaCl_(2)` (0.36 mole) to `100 mL` of `H_(2)O` raises the temperature from `20^(@)C` to `90^(@)C`. Similarly, when `30 g NH_(4)NO_(3)` (0.3 mole) are dissolved in `100 mL` of water at `20^(@)C`, the temperature will be lowered to `0^(@)C`. A typical hot or cold pack works for about 20 min. |
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| 857. |
The given graph shows the vapour pressure - temperature curves for some liquids Liquids A,B, C and D respectively areA. diethyl ether, acetone , ethyl alcohol , waterB. acetone , ethyl alcohol, diethyl ether , waterC. water, ethyl alcohol , acetone , diethyl etherD. ethyl alcohol, acetone , diethyl ether, water |
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Answer» Correct Answer - A The vapour pressure increases with decrease in intermolecular forces . When the forces are weak, the liquid has high volatility and maximum vapour pressure . Diethyl ether has highest vapour pressure while water has lowest vapour pressure. |
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| 858. |
In three beakers labelled as (A),(B) and (C ) , 100 mL of water, 100 mL of 1 M solution of glucose in water and 100 mL of 0.5 M solution of glucose in water are taken respectively and kept at same temperature . Which of the following statements is correct ?A. Vapour pressure in all the three beakers is sameB. Vapour pressure of beaker B is highestC. Vapour pressure of beaker C is highestD. Vapour pressure of beaker B is lower than that of C and vapour pressure of beaker C is lower than that of A . |
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Answer» Correct Answer - D Vapour pressure of a solution containing non-volatile solute is less than that of the pure solvent . The decrease in vapour pressure depends upon the quantity of non-volatile solute present in it. Hence, vapour pressure of A gt C gt B . |
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| 859. |
A 0.0020 m aqueous solution os an ionic compound `[Co(NH_(3))_(5)(NO_(2))]` CI freezes at `-0.0732^(@)C`. Number of moles of ions which 1 mole of an ionic compound produces on being dissolved in water will be:A. 3B. 4C. 1D. 2 |
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Answer» Correct Answer - d `DeltaT_(f)=i K_(f)m` `i=((0.0732K))/((1.86Km^(-1))xx(0.0020m))=2` Since, the ionic compound complete dissociates in aqueous olution, degree of dissociation `(prop)is ~~1`. For dissociation, `alpha=(i-1)/(n-1)` `1=(2-1)/(n-1)or n=1+1=2` No. of mole ions formed=2 |
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| 860. |
Define the term colligative property. Give examples. |
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Answer» (1) Colligative Property : The property of a solution which depends on the total number of particles of the solute (molecules, ions) present in the solution and does not depend on the nature or chemical composition of solute particles is called colligative property of the solution. (2) Examples of colligative properties : (a) lowering or relative lowering of vapour pressure of a solution (b) elevation in the boiling point (c) depression in the freezing point (d) osmotic pressure. |
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| 861. |
Explain the following terms : (1) Relative vapour pressure of a solution (2) Lowering of vapour pressure of a solution (3) Relative lowering of vapour pressure. |
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Answer» (1) Relative vapour pressure of a solution : If P0 is the vapour pressure of a pure liquid (solvent) and P is the vapour pressure of a solution after adding a nonvolatile solute, then, relative vapour pressure = \(\frac{P}{P_0}\). (2) Lowering of vapour pressure of a solution : When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure. If P0 is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P0, then, (P0 – P) is the lowering of the vapour pressure. (3) Relative lowering of vapour pressure : If P0 and P are the respective vapour pressures of a pure liquid (solvent) and the solution containing a non-volatile solute then P < P0. Hence, P0 – P represents the lowering of the vapour pressure due to addition of a nonvolatile solute. ∴ Relative lowering of vapour pressure = \(\frac{P_0-P}{P_0}\) = \(\frac{ΔP}{P_0}\) |
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| 862. |
What is the mole fraction of glucose in 10% w/W glucose solution ?A. 0.01B. 0.02C. 0.03D. 0.04 |
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Answer» Correct Answer - A No. of moles of glucose =`10/180`=0.0555 mol No. of moles of water =`90/18`=5 mol Number of moles of solution=5.0555 mol Mole fraction of glucose =`"No. of moles of glucose"/"No. of moles of solution"=0.0555/5.0555=0.01` |
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| 863. |
Calculate the percentage composition of a solution obtained by mixing 200 g of a `20%` and 300 g of a `30%` solution by weight.A. `50%`B. `28%`C. `64%`D. `24%` |
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Answer» Correct Answer - D Solute in 200 g of 30% solution =60 g Solute in 300 g of 20% solution = 60 g Total grams of solute = 120 g Total grams of solution = 200 +300=500 g % of solute in the final solution=`120/500xx100`=24% |
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| 864. |
Explain reverse osmosis. |
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Answer» 1. The pure water moves through the semipermeable membrane to the NaCl solution due to osmosis. 2. This process can be reversed by applying pressure greater than the osmotic pressure to the solution side. Now the pure water moves from the solution side to the solvent side and this process is called reverse osmosis. 3. Reverse osmosis can be defined as a process in which a solvent passes through a semipermeable membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater than the osmotic pressure. |
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| 865. |
The degree of dissociation α is equal to … |
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Answer» The degree of dissociation α is equal to \(\frac{i-1}{n-1}\). |
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| 866. |
Define the term bisotonic. |
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Answer» 1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. 2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e., the net solvent flow between two isotonic solutions is zero. |
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| 867. |
You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one super saturated. How would you determine each solution? |
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Answer» 1. Saturated solution: When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution. 2. Unsaturated solution: When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution. 3. Super saturated solution: It is a solution that holds more solute than it normally could in its saturated form. Example: 1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCI in 1 liter of water at 25°C. 2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCI in 1 litre of water at 25°C. 3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCI in 1 liter of water at 25°C. |
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| 868. |
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood. What is the osmotic pressure of blood? |
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Answer» C = 0.25 M T = 370.28 K (π)glucose = CRT (π) = (0.25 mol L-1) x (0.082 L atm K-1 morl-1) x (370.28 K) = 7.59 atm |
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| 869. |
Number of moles of the solute dissolved per dm3 of solution is …… |
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Answer» Number of moles of the solute dissolved per dm3 of solution is molarity. |
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| 870. |
Define solution with an example. |
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Answer» 1. A solution is a homogeneous mixture of two or more substances consisting of atoms. ions or molecules. 2. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained. In this solution, Na+ and C- ions are uniformly distributed in water. Here NaCI is the solute and water is the solvent. |
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| 871. |
What are aqueous and non aqueous solution? Give example. |
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Answer» 1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution. Example, salt in water. 2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl4 etc, the resultant solution is called a non aqueous solution. Example, Br, in CCI4. |
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| 872. |
Define osmotic pressure. |
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Answer» Osmotic pressure can be defined as the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semipermeable membrane. |
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| 873. |
State and explain Henry’s law. |
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Answer» Henry’s law: This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.” Psolute ∝ xsolute in solution Psolute = KH . xsolute in soluation xsolute = mole fraction of solute in the solution KH = empirical constant. Psolute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of KH depends on the nature of the gaseous solute and solvent. |
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| 874. |
When NaCl is dissolved in water, boiling point …… |
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Answer» When NaCl is dissolved in water, boiling point increases. |
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| 875. |
What are the limitations of Henry’s law? |
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Answer» 1. Henry’s law is applicable at moderate temperature and pressure only. 2. Only the less soLuble gases obey Henry’s law. 3. The gases reacting with solvent do not obey Henry’s law. 4. The gases obeying Henrys law should not associated or dissociated while dissolving in the solvent. |
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| 876. |
Define forniality. |
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Answer» Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution. Formality (F) = (Number of formula weight of solute) / (Volume of the solution (1 L)) |
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| 877. |
Out of `M H_(2)SO_(4)` and `1H J_(2)SO_(4)` which is mor econcentreated and why ? |
| Answer» `1m H_(2)SO_(4)` is more concerntrated. | |
| 878. |
A sample of drinking water was found to be severely contaminated with chloroform, `CHCl_(3)`, supposed to be carcinogen. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.A. `2.317xx10^(-4)m`B. `0.7864xx10^(-4)`C. `1.255xx10^(-4)m`D. `0.5555xx10^(-4)m` |
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Answer» Correct Answer - 3 15 ppm means `15 g` of `CHCl_(3)` in `10^(6) g (or 1000 kg)` of solution. Since `CHCl_(3)` is present in trace amounts we can (approximately) consider mass of solvent equal to the mass of solvent, i.e., `1000 kg H_(2)O` contains `15 g CHCl_(3) ("or "15 g//119.5 g mol^(-1)=0.1255 mol CHCl_(3))`. Thus, 1 kg of the sample will contain. `0.1255/1000 mol CHCl_(3)` `=1.255xx10^(-4) mol CHCl_(3)` Thus, the molality of the `CHCl_(3)` in the aqueous solution is `1.255xx10^(-4) m`. |
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| 879. |
Of `0.1` molal solutions of glucose and sodium chloride respectively. Which one will have a higher boiling point? |
| Answer» Correct Answer - `0.1` molal `NaCl` solution will have higher boiling point. | |
| 880. |
When mercuric iodide is added to the aqueous solution of potassium iodide, then:A. freezing point id raisedB. freezing point is loweredC. freezing point does not changeD. boiling point is raised |
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Answer» Correct Answer - A `2KI+HgI_(2)rarrK_(2)[HgI_(4)]` Number of particles decreases due to above reaction hence freezing point is raised. |
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| 881. |
A molal solution is one that contains one mole of a solute inA. 1000 g of the solventB. one litre of solventC. one litre of solutionD. 22.4 litre of the solution |
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Answer» Correct Answer - A A one molal solution contains one mole of solute in 1000 g of solvent. |
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| 882. |
Out of 0.1 molal solution of glucose and sodium chloride respctively, which one will have higher boiling point ? |
| Answer» 0.1 molal solution of sodium chloride will have higher `DeltaT_(b)` and higher boiling point as well because it dissociates into ions. At the sametime, 0.1 molal glucose solution being a molecular solid will not dissociate into ions. | |
| 883. |
An ideal solution is formed when its components sameA. high solubilityB. no enthalpy change on mixingC. no volume change on mixingD. baton (2) and (3) |
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Answer» Correct Answer - 4 An ideal solution satisfies the following conditions: (i) There will be no change in volume on mixing the two components, i.e., `DeltaV_("mixing")=0` (ii) There will be no change in enthalpy (i.e no heat energy is absorbed or releases) when the two components are mixed, i.e., `DeltaH_(mix)=0)`. |
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| 884. |
How are `DeltaT_(b)` and `DelatT_(f)` related to the molar mass of the solute ? |
| Answer» Both `DeltaT_(b) and DelatT_(f)` are inversaly proportional to the molar mass of the solute. | |
| 885. |
Find the molaity and molatity of a 15% solution w/w of `H_(2)SO_(4)("density of "H_(2)SO_(4)=1.02gcm^(-3))`. |
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Answer» Calculation of molality of the slution, Weight of `H_(2)SO_(4)` in 100 g of solution=15.0 g Molar mass of `H_(2)SO_(4)=98 g mol^(-1)` Molality of solution (m) `=("No.of gram moles of "H_(2)SO_(4))/("Mass of warter in kg")=(15g//98gmol^(-1))/(0.085 kg)` `= 1.8(mol kg^(-1))=1.8m" Calculation of molarity of the solution. Molarity of solution (M)" = ("No. of gram moles moles of "H_(2)SO_(4))/("Volume of solution in litres")` `"Weight of solution "=("Weight of solution")/("Density")=((100g))/((1.02g cm^(-3))=98.0cm^(3)` No. of gram moles of `H_(2)SO_(4)=((15.0g))/((98gmol^(-1)))=0.153 mol` `"Molarity of solution (M)"= ((0.153mol))/(((98.0)/1000dm^(3)))=1.56mol (dm^(3))^(-1)=1.56 M` |
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| 886. |
Marine life like fish prefers to stay at lower level in water. Explain. ORExplain, why do aquatic animals prefer to stay at lower level of water during summer? |
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Answer» 1. The solubility of oxygen gas decreases with the increase in temperature. 2. In sea or lake water, the temperature of upper level is higher than the lower level. 3. Therefore the dissolved oxygen content in water is more at lower level than at higher level required for the marine life. Hence marine life like fish prefers to stay at lower level than upper level of water. |
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| 887. |
At 25°C, 3g of a solute A in 100 ml of an aqueous solution gave an osmotic pressure of 2.5 atm. What is the nature of the solute (associated or dissociated) if its normal molar mass is 246? |
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Answer» Л = i x C x R x T 2.5 = i x ( 3/246) x ( 0.0821 / 0.1) x 298 i = 2.01 As i > 1 the solute A undergoes dissociation. |
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| 888. |
A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at `25^(@)"C"`. Which of the following operation result will not in a decrease in the equilibrium vapour pressure at `25^(@)"C"`?A. Moving the piston downward a short distanceB. Removing a small amount of vapourC. Removing a small amount of liquid waterD. Dissolving some salt in the water. |
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Answer» Correct Answer - A,B,C, Solution: Colligative property |
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| 889. |
The vapour pressure of pure water at `26^(@)"C"` is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g glucose,`"C"_(6)"H"_(12)"O"_(6)`, in 70 g water? |
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Answer» Correct Answer - 24.5 torr Glucose-B, water-A `(P_(A)^(0)-P_(sol))/(P_(A)^(0))=(W_(B)xxm_(A))/(m_(B)xxw_(A))` `(25.21-P_(sol))/25.21=(20xx18)/(180xx70)` `P_(sol)=24.5 " torr"` |
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| 890. |
For A and B to form an ideal solution which of the following conditions should be satisfied?A. `DeltaH_("mixing")=0`B. `DeltaV_("mixing")=0`C. `DeltaS_("mixing")=0`D. All three conditions mentioned above. |
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Answer» Correct Answer - D In ideal solution intermolecular forces among molecules of A and B are identical to that among the molecules of A and to that anong the molecules of B. `:.DeltaH_("mixing")=0` `DeltaV_("mixing ")=0` `DeltaS_("mixing")=0` |
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| 891. |
The mole fraction of toluene in vapour phase which is in equlibrium with a solution of benzene and toluene having a mole fraction of toluene `0.500` is (vapour pressure of pure benzene and pure toluene are 119 torr and 37.0 torr respectively at the same temperature).A. `0.5`B. `0.763`C. `0.237`D. `1` |
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Answer» Correct Answer - C `P=P_(B)^(0)X_(B)+P_(T)^(0)X_(T)` `p= 119xx0.5+37xx0.5` `Y_(T)=(P_(T)^(0)X_(T))/(P)=(37xx0.5)/(119xx0.5+37xx0.5)=0.237` |
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| 892. |
Calculate the mole fraction of toluence in the vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50. The vapour pressure of pure benzene is 199 torr, that of toluene is 37 torr at the same temperature. |
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Answer» Correct Answer - 0.237 `C_(6)H_(6)"-A",C_(7)H_(8)"-B"` `X_("B")=0.50,X_("A")=1-0.5` =0.5 `Pt=X_("A")P_("B")^(0)+X_("B")P_("B")^(0),P_("B")=X_("B")P_("B")^(0)` `Y_("B")=(P_("B"))/(Pt)` `(X_("B")P_("B")^(0))/(X_("A")P_("A")^(0)+X_("B")P_("B")^(0))=(0.5xx37)/(0.5xx119+0.5xx37)=37/156=0.237` |
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| 893. |
Which one of the following given below concerning properties of solutions, describe a colligative effect ?A. Boiling point of pure water decreases by the addition of ethanolB. Vapour pressure of pure water decreases by the addition of nitric acidC. Vapour pressure of pure benzenen decreases by the addition of napththaleneD. Boiling point of pure benzene increases by the addition of toluene |
| Answer» When solute added to the solvent, the vapour pressure of solution decreases. | |
| 894. |
The molality of a soluton having 18 g of glucose (Molar mass=180) dissolved in 500 g of water is :A. 1.2mB. 0.4mC. 0.1mD. 0.2m |
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Answer» Correct Answer - D `m = (18 xx 1000)/(180 xx 500) = 0.2m` |
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| 895. |
A solution has 1:4 mole ratio of pentane to bexane. The vapour pressure of the pure hydrocarbhonse at `20^(@)C` and 440 mm Hg is for pentane and 120 mm Hg for bexane: The mole fraction of pentane in the vapour phase will be :A. 0.2B. 0.549C. 0.786D. 0.478 |
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Answer» Correct Answer - c `(n_(C_(5)H_(12)))/(n_(C_(6)H_(14)))=1/4` `X_(C_(5)H_(12))=1/(1+4)=1/5` `X_(C_(6)H_(14))=4/(1+4)=4/5` `P_(C_(5)H_(12))^(@)=440mmHg,P_(C_(6)H_(14))^(@)=120 mm Hg` `P_(("total"))=P_(C_(5)H_(12))^(@)xxX_(C_(5)H_(12))+P_(C_(6)H_(14))^(@)xxX_(C_(6)H_(14))` `=440xx1//5+120xx4//5` `=88+96=184 mm Hg` Mole fraction `C_(5)H_(12)` in vapour phase `X_(C_(5)H_(12))=("Partoa V.P. of "C_(5)H_(12))/("Partoa V.P. of "C_(6)H_(14))=(88mm)/(184mm)=0.478` |
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| 896. |
How do students start to Practice the Solutions Class 12 Mock Test? |
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Answer» Practice the Solutions Class 12 Online Mock Test. This MCQ Questions of solution chapter test is as per the latest Format of the CBSE Board examination. Free Solutions Class 12 Online Mock Test for Chemistry for important topics of all chapters in Class 12 Chemistry book. Access full online mock test Chemistry Class 12 that desgined based on latest pattern of questions. Our teachers created test series as per latest syllabus and exam pattern. You should practice these Chemistry Class 12 Online Practice Tests having objective-based questions and compare your answers with the solutions provided by us. This Class 12 Online Mock Test Solutions are more helpful for students to score high marks in exams and clear all doubts of this chapter. Before your exam, take our online test series to know where you stand and develop exam strategies and learn time management by solving class 12 Solutions Mock Test. Click here to start practice: - Class 12 Science Solutions Mock Test |
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| 897. |
Some entropy change are represented infigure. Select correct entropy change. A. `DeltaS_(1),DeltaS_(2),DeltaS_(3)`B. `DeltaS_(1),DeltaS_(2),DeltaS_(4)`C. `DeltaS_(1),DeltaS_(2),Delta_(3),DeltaS_(4)`D. `DeltaS_(2)` and `DeltaS_(4)` |
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Answer» Correct Answer - C All `DeltaS_(1),DeltaS_(2),DeltaS_(4)` are correct entropy changes. |
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| 898. |
What happens to the net vapour pressure of resultant solution- it increases or decreases or remains same? Support your answer with suitable explanation. |
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Answer» The net vapour pressure of the solution decreases because newly formed hydrogen bonds are stronger forces of attraction than the original forces of attraction existing in separate solutions. |
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| 899. |
Define binary solution? |
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Answer» Binary solution is a solution containing only one solute dissolved in a solvent. |
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| 900. |
What is molarity? |
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Answer» The number of moles of solute dissolved in one litre or 1dm3 of solution is known as molarity. |
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