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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.
| 751. |
The concentration term is used in the calculation of vapour pressure of solution is …… |
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Answer» The concentration term is used in the calculation of vapour pressure of solution is Mole fraction. |
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| 752. |
The concentration term used in the neutralisation reactions is …. |
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Answer» The concentration term used in the neutralisation reactions is Normality. |
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| 753. |
What are liquid solutions ? Explain with example. |
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| 754. |
How will you prepare a standard solution? |
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Answer» 1. A standard solution or a stock solution is a solution whose concentration is accurately known. 2. A standard solution of required concentration can be prepared by dissolving a required amount of a solute in a suitable amount of solvent. 3. It is done by transforming a known amount of solute to a standard flask of definite volume. A small amount of water is added lo the flask and shaken well to dissolve the salt. 4. Then water is added to the flask to bring the solution level lo the mark indicated at the top end of the flask. 5. The flask is stoppered and shaken well to make concentration uniform. |
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| 755. |
Match the column I with column II.column Icolumn IIA. Gas in gas(i) unsaturatedB. NaCl(ii) unstableC. Fat(iii) He - O2D. Super saturated solutions(iv) 36 gE. Dilute(v) ether |
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Answer» A. (iii) B. (iv) C. (v) D. (ii) E. (i) |
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| 756. |
Carbonated water is an example for …… |
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Answer» Carbonated water is an example for Liquid solution |
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| 757. |
The concentration of commercially available H2O2 is … |
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Answer» The concentration of commercially available H2O2 is 3% |
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| 758. |
Humid oxygen is an example of …… |
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Answer» Humid oxygen is an example of Gaseous solution. |
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| 759. |
What are solid solution? Give example. |
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| 760. |
Suppose a solid solution is formed are very small. What kind of solid solution is this likely to be? |
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Answer» Solid in solid type. E.g: Copper in gold. This type of solutions are called alloys. |
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| 761. |
A solution of glucose in water is labelled as `10%` w/w. What would be the molarity of the solution ? |
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Answer» Given `10%((W)/(w))` glucose in water solution Weight of glucose = 10 gms GMW of glucose `(C_(6)H_(12)O_(6))=180` Weight of water `=100-10=90" gms"` Molarity (m) `=("Weight")/("GMW")xx(1000)/(W("gms"))` ltbbrgt `=(10)/(180)xx(1000)/(90)=(100)/(18xx9)` `=0.617" moles"//"kg"` Molarity `=("Weight")/("GMW")xx(1000)/("V"(ml))` Weight of solution = 100 gms Assuming that Density of solution = 1.2 gm/ml `:." Volume "=(100)/(1.2)=83.33" ml"` `:." Molarity (M)"=(10)/(180)xx(1000)/(83.33)"ml"` `=(1000)/(18xx83.33)=0.67" M"` |
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| 762. |
An example of solid homogeneous mixture is … |
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Answer» An example of solid homogeneous mixture is Brass |
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| 763. |
A mixture of N2, O2, CO2 and other traces of gases is known as ……… |
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Answer» A mixture of N2, O2, CO2 and other traces of gases is known as Air |
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| 764. |
..... is used for dental filling. |
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Answer» Amalgam of potassium is used for dental filling. |
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| 765. |
Define solubilitv. |
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Answer» The solubility of a substance is defined as the amount of the solute that can be dissolved in loo g of the solvent at a given temperature to form a saturated solution. |
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| 766. |
Ammonia is more soluble than oxygen in water. Why? |
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Answer» Ammonia forms hydrogen bonding with water molecules, this intermolecular bonds arc very strong and thus the ammonia is more soluble in water. Ammonia is strongly interact with water to form ammonium hydroxide. But oxygen is more electronegative it is not able to interact with water more. So NH3 is more soluble than O2 in water. |
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| 767. |
Concentrated nitric acid used in the laboratory work is `68%` nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the denisty of the solution is `1.504" mL"^(-1)` ? |
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Answer» Given `68%" HNO"_(3)` by mass that means 68gms mass of `"HNO"_(3)` present in 100 gms of solution. Molecular weight of `"HNO"_(3)=63` Number of moles of `"HNO"_(3)=("weight")/("GMW")=(68)/(63)=1.079` Given density of the solution = 1.504 gm/ml Volume of solution `= ("Mass of solution")/("density")` `=(100)/(1.504)=66.5" ml"` Molarity `=nxx(1000)/("V"(ml))` `=(1.079xx1000)/(66.5)=16.23" M"` |
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| 768. |
When acetone and chloroform are mixed together , hydrogen bonds are formed between them. Which of the following statements is correct about the solution made by mixing acetone and chloroform ?A. On mixing acetone and chloroform will form an ideal solutionB. On mixing acetone and chloroform positive deviation is shown since the vapour pressure increasesC. On mixing acetone and chloroform negative deviation is shown since there is decreae in vapour pressureD. At a specific composition acetone and chloroform will form minimum boiling azeotrope |
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Answer» Correct Answer - C When acetone and chloroform are mixed together, hydrogen bonds are formed between them which increases intermolecular interactions hence decreases the vapour pressure showing negative deviation . |
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| 769. |
The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is (a) 24 mm Hg (b) 32 mm Hg (c) 48 mm Hg (d) 12 mm Hg |
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Answer» Correct answer is (d) 12 mm Hg |
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| 770. |
Cryoscopic constant depends on(a) nature of solvent (b) nature of solute (c) nature of solution (d) number of solvent molecules |
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Answer» (a) nature of solvent Cryoscopic constant depends on nature of solvent. |
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| 771. |
The colligative property of a solution is (a) vapour pressure (b) boiling point (c) osmotic pressure (d) freezing point |
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Answer» (c) osmotic pressure The colligative property of a solution is osmotic pressure. |
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| 772. |
Which one is a colligative propertyA. Boiling pointB. Vapour pressureC. Osmotic pressureD. Freezing point |
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Answer» Correct Answer - C Osmotic pressure is an example of colligative property because its value depends only on the number of moles of solute, not on their chemical nature. |
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| 773. |
Calculate the volume of 80% (by mass) of `H_(2)SO_(4)` (density=1.8 g/mL) reqiored tp [reare 1 litre of 0.2 M `H_(2)SO_(4)`. |
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Answer» Correct Answer - 13.61 mL Step I. Calculation of mass of `H_(2)SO_(4)` `"Molarity of acid (M)"=0.2 M=0.2"mol L"^(-1), "Volume os solution"=1 L` Molar mass of acid = 98 g `mol^(-1)` `"Molarity"=("Mass/Molar mass")/("Volume of solution in Litres")` `(0.2" mol L"^(-1))=W/((98" g mol"^(-1))xx(1L))` `1=(0.2" mol L"^(-1))xx(98" g mol"^(-1))(1L)=19.6 g`. step II. Calculation of volume os 80% `H_(2)SO_(2)` `"Volume of pure"H_(2)SO_(4)=("Mass")/("density")=((19.6g))/((1.8" g mol"^(-1)))=10.89 mL` `"Volume of 80% "H_(2)SO_(4)=(10.89mL)xx100/80=13.61 mL` |
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| 774. |
Determine the molaarity of an antifreeze soliution sontaining 250 g water mixed with 222 g of ethylene glylene glycol `(HOCH_(2)CH_(2)OH)`. The dnesity of solution is 1.07 g/mL. |
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Answer» Correct Answer - 8.12 M Mass of ethylene glycol = 222 g Mass of water = 250 g Mass of solution = (222+250)=472 g Density of solution = 1.07 g `mL^(-1)` `"Volume of solution (M) "=("Mass os ethylene glycol/molar mass")/("Volume os solution in litres")` `=((222g))/((62" g mol"^(-1))xx(0.4412L))` =8.12 mol `L^(-1)`=8.12 M |
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| 775. |
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution. |
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Answer» 300 g of 25% solutions contains solute = 75 g 400 g of 40% solution contains solute = 160 g. Total solute = 160 + 75 = 235 g Total solution=300 + 400 = 700 g % of solute in final solution = \(\frac{235}{700}\) × 100 = 33.5% % of water in the final solution = 100 - 33.5 = 66.5% |
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| 776. |
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample. |
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Answer» 15 ppm means 15 parts in million (106) parts by mass in solution ∴ % by mass = \(\frac{15}{10^6}\) × 100 = 15 × 10-4 Taking 15 g chloroform in 106g Molar mass of CHCl3 = 12 + 1 + 3 × 35.5 = 119.5 g mol-1 Molality = \(\frac{\frac{15}{119.5}}{10^6}\) × 1000 = 1.25 × 10-4 m |
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| 777. |
Calculate the molarity of each of the following solutions: 30g of `Co(NO_(3))_(2)6H_(2)O` in 4.3L of solution `30 mL of0.5 M H_(2)SO_(4)` diluted to 500 mL |
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Answer» `"Molarity of solution"=("Mass of solute/Molar mass of solute")/("Volume of solution in litres")` Molar mass solute, `Co(NO_(3))_(2).6H_(2)O=30 g`. Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=219 g mol^(-1)` Volume of solution=4.3 L `"Molarity(M)"=((30g)//(291g mol^(-1)))/((4.3L))=0.024 molL^(-1)=0.024 M` Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))`=30 mL Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))`=1.5 M Volume of diluted `H_(2)SO_(4)` solution `(V_(2))`=500 mL Molarity of diluted `H_(2)SO_(4)(M_(2))` can be calculated as : `M_(1)V_(1)=M_(2)V_(2)` `M_(2)=(M_(1)V_(1))/V_(2)=((30mL)xx(0.5M))/((500mL))=0.03 M` |
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| 778. |
The solution in which the blood cells remain their normal shape, with regard to the blood, areA. IsotonicB. HypertonicC. HypotonicD. None of these |
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Answer» Correct Answer - A Osmosis will not take place. |
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| 779. |
For a solvent `Delta_(freez)H^(@)=5355" cal mole"^(-1)`, molar mass is 150 and freezing point is `80^(@)C` is `80^(@)C`, its `K_(f)` will beA. `3.6" K kg mole"^(-1)`B. `0.12" K kg mole"^(-1)`C. `0.18 K kg mole"^(-1)`D. `4.8 K kg mole"^(-1)` |
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Answer» Correct Answer - 4 |
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| 780. |
If liquids A and B form an ideal solution.A. The free energy of mixing is zeroB. The free energy as well as the entropy of mixing are each zeroC. The enthalpy of mixing is zeroD. The entropy of mixing is zero |
| Answer» Correct Answer - C | |
| 781. |
Which liquids form ideal solution? |
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Answer» Liquids having similar structure and polarities. |
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| 782. |
Which substance is usually added into water in the car radiator to act as antifreeze? |
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Answer» Ethylene glycol is usually added into water in the car radiator to act as antifreeze. |
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| 783. |
0.1 M NaCI and 0.1 M `CH_(3)COOH` solution are kept in seprate containers. If their osmotic pressures are `P_(1)"and"P_(2)` repectively then what will be the correct statement?A. `P_(2)gtP_(2)`B. `P_(2)=P_(2)`C. `P_(2)ltP_(2)`D. `P_(2)=P_(2)=0 atm, |
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Answer» Correct Answer - a Since NaCI is a stronger electrolyte than `CH_(3)COOH`, it will dissociate to freater extent, and release ion (Particle). Therefore, `P_(1)gtP_(2)`. |
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| 784. |
If all the following four compounds were sold at th same price, which would be the cheapest for preparing antifreeze solution for a car radiator ?A. `CH_(3)OH`B. `C_(2)H_(5)OH)`C. `C_(2)H_(4)(OH)_(2)`D. `C_(3)H_(5)(OH)_(3)` |
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Answer» Correct Answer - a `DeltaT_(f)prop1/M_(B)` `DT_(f)` will be maximum in case of `CH_(3)OH` because its molar mass is sthe least. Therefore, it will be the cheapest. |
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| 785. |
Antifreeze are the substances whichA. Stop freezingB. Decreases free pointC. Increases freezing pointD. Melt the ice |
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Answer» Correct Answer - 2 |
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| 786. |
Which among the following has highest boiling point ?A. 1 M glucoseB. 1 M KClC. 1 M Al`(NO_(3))_(3)`D. 1 M `Na_(2)SO_(4)` |
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Answer» Correct Answer - 3 |
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| 787. |
On freezing dilute aq. NaCl which crystallise out firstA. Both togetherB. WaterC. NaClD. (NaOH+HCl) |
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Answer» Correct Answer - 2 |
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| 788. |
What should be the mass of glucose to be added to 900 g water to decrease vapour pressure by 1% ?A. 181.8 gB. 90.9 gC. 46 gD. 136.36 g |
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Answer» Correct Answer - 2 |
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| 789. |
Correct increasing order of osmotic pressure for the following isA. Sucrose (0.1 M) `lt` glucose (0.5 M)`lt `urea(1M) `lt` NaCl(2M)B. Glucose(0.5 M)`lt` urea (1M)` lt `NaCl(2M)`lt`Sucrose(0.1 M)`C. Urea(1 M)`lt`NaCl(2M)`lt`glucose(0.5 M)`lt Sucrose(0.1 M)`D. NaCl(2M)`lt` sucrose(0.1 M)`lt` glucose(0.5 M)`lt `urea(1M) |
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Answer» Correct Answer - 1 |
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| 790. |
In 100 g naphthalene 2.56 g sulphur is added boiling point of solution (solid) decreases by `0.68^(@)C`, atomicity of sulphur is (`K_(f)` of naphthalene=6.8 `Km^(-1)`)A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - 4 |
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| 791. |
Which of the following solution has higher freezing point ? Justify your answer. `0.05 M AI_(2)(SO_(4))_(3)` `0.1 M K_(3)[Fe(CN)_(6)].` |
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Answer» `DeltaT_(f)prop"No. of particles=i"xx"concerntration"` `"For 0.05 M" AI_(2)(SO_(4))_(3)"solution,"DeltaT_(f)=5xx0.05=0.25" mole ions."` `"For 0.1 M "K_(3)(Fe(CN)_(6))] "solution," DeltaT_(f)=4xx0.1=0.4 "mole ions."` Depression in freezing point temperature for `AL_(2)(SO_(4))_(3)` solution in less, It has therefore, higher freezing point temperature. |
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| 792. |
What do you mean by relative lowering in vapour pressure ? |
| Answer» For answer consult section 9. | |
| 793. |
Define Hypertonic solution. Explain why depressionin freezing point is a colligative property. Calculate the molar mass of solute with it. |
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Answer» A solution is said to be hypertonic if its osmotic pressure is mor etan the of other separated by a semipermeable membrane. For answer consult section 11. |
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| 794. |
Which colligative property is generally used for determining the moar mass of a solute ? |
| Answer» Osmotic pressure is generally used for this purpose. | |
| 795. |
What is the purpose of adding ethlene glycol to water ? |
| Answer» Ethylene glycol `(CH_(2)OHCH_(2)OH)` is added to water to lower its freezing point. It is known as antifreeze solution. | |
| 796. |
What is the effect of temperature on molarity of a solution ? |
| Answer» It changes with change in temperature. | |
| 797. |
why does the solubility of NaCI in water increses with the rise in temperature ? |
| Answer» Because the p[rocess of dissolution is of endothermic nature. | |
| 798. |
`1.0` molal aqueous solution of an electrolyte `X_(3)Y_(2)` is 25 % ionized. The boilling point of solution is (`K_(b)` for `H_(2)O=0.52K kg//mol`)A. `375.5K`B. `374.04K`C. `377.12K`D. `373.25K` |
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Answer» Correct Answer - B `underset(1-alpha)(x_(3)y_(2))hArr underset(n alpha)(3x^(2+))+underset(m alpha)(2y^(3-)) "for complete ionization"` `i=1+(m+n-1)alpha` `i=1+(2+3-1)xx0.25=1+1=2` `DeltaT_(b)=ixxkbxxm=2xx0.52xx1=1.04` B.P. of solution`(T_(b))=DeltaT_(b)+T_(b)^(@)=1.04+373=374.04K` |
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| 799. |
Out of 1 M glucose and 2M glucose which has a higher b pt and why? |
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Answer» 2M glucose will have higher boiling point because boiling point of a solution of a nonvolatile liquid increases with increase in concentration. |
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| 800. |
What happens when the external pressure applied becomes more than the osmotic pressure of solution? |
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Answer» When external pressure applied is more than the osmotic pressure then reverse osmosis takes place. The solvent will then flow from the solution to the five solvent by semipermeable membrane. |
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