InterviewSolution
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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.
| 1. |
10mL of 2(M) NaOH solution is added to 200mL of 0.5 (M) of NaOH solution. What is the final concentration?A. 0.57MB. 5.7MC. 11.4MD. 1.14M |
| Answer» Correct Answer - A | |
| 2. |
Dissolving 120 g of urea (molecular mass = 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is :A. 1.78MB. 2.00MC. 2.05MD. 2.22M |
| Answer» Correct Answer - C | |
| 3. |
The density of a solution prepared by dissolving 120 g of urea (mol. Mass=60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution isA. 2.05 MB. 0.50MC. 1.78MD. 1.02M |
| Answer» Correct Answer - A | |
| 4. |
The incorrect statement (s) regarding 2M `MhCl_(2)` aqueous solution is/are `(d_("solution") = 1.09 gm//ml)`A. Molality of `Cl^(-)` is 4.44 mB. Mole fraction of `MgCl_(2)` is exactly 0.035C. The conc. Of `MgCl_(2)` is 19% w/vD. The conc. of `MgCl_(2) " is " 19 xx 10^(4)` ppm |
| Answer» Correct Answer - B::D | |
| 5. |
Find number of `Na^(+)` & `PO_(4)^(-3)` ions in `250mol` of `0.2M Na_(3)PO_(4)` solution . |
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Answer» `Na_(3)PO_(4) + aq. Rarr 3 Na^(+) (aq) + PO_(4)^(-3)` (aq) [Ionic compound when added to water ionize compleltely] 50 millimoles (m.m) 150 mm 50 mm No. of `Na^(+) " ions " = 150 xx 10^(-3) xx N_(A)` , No. of `PO_(4)^(-3) " ions " 50 xx 10^(-3) xx N_(A)` |
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| 6. |
`100 ml "of" 0.06 M Ca (NO_(3))_(2)` is added to `50 mL` of `0.06 M Na_(2)C_(2) O_(4)`. After the reaction is complete.A. 0.003 moles of calcium oxalate will get precipitatedB. 0.003 M `Ca^(2+)` will remain in excessC. `Na_(2)C_(2)O_(4)` is the limiting reagentD. Oxalate ion `(C_(2)O_(4)^(2))` concentration in final solution is 0.003 M |
| Answer» Correct Answer - A::C | |
| 7. |
Molarity of liquid HCl with density equal to `1.17 g//mL` is:A. 36.5 MB. 18.25 MC. 32.05 MD. 42.10 M |
| Answer» Correct Answer - C | |
| 8. |
Molality of 20% (w/w) aq. glucose solution isA. `(25)/(18)m`B. `(10)/(9)m`C. `(25)/(9)m`D. `(5)/(18)m` |
| Answer» Correct Answer - A | |
| 9. |
What would be the molality of 20% (mass/mass) aqueous solution of KI ? (molar mass of KI `= 166 g mol^(-1)`)A. `1.08`B. `1.48`C. `1.51`D. `1.35` |
| Answer» Correct Answer - C | |
| 10. |
The molar concentration of HCl (aq.) is `10^(-5)`M. Which of the following statement are correct (`d_("solution")` = 1 gm/cc)A. The mole fraction of `HCl ~= 1.8 xx 10^(-7)`B. The concentration of HCl in ppm is 3.65 ppmC. The molality of HCl solution is approximately `10^(-5) m`D. The (w/v)% of solution is `3.65 xx 10^(-5) %` |
| Answer» Correct Answer - A::C::D | |
| 11. |
What volume of `1M` & `2M H_(2)SO_(4)` solution are required to produce `2L` of `1.75M H_(2)SO_(4)` solution ? . |
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Answer» Let XL be vol. of 1M solution. `:. (2 - X)L` is vol. of 2M solution Moles of `H_(2)SO_(4) = 2 xx 1.75 = 1 (X) + (2 - X) 2` `3.5 = 4 - X , X = 0.5 L` i.e., 0.5 L of 1 M & 1.5 L of 2M solution requires |
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| 12. |
Statement-1 : Molality of pure ethanol is lesser than pure water Statement-2 : As density of ethanol is lesser than density of water [Given : `d_("ethanol") = 0.789 gm//ml, d_("water") = 1gm//ml`]A. Statement-1 is true, statement -2 is true and statement-2 is correct explanation for statement-1B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1C. Statement-1 is false, statement-2 is trueD. Statement-1 is true, statement-2 is false |
| Answer» Correct Answer - B | |
| 13. |
The molarity of a solution of sodium chloride (mole wt. = 50.5) in water containg 5.85 gm of sodium chloride in 500 ml of solution isA. `0.25`B. `2.0`C. `1.0`D. `0.2` |
| Answer» Correct Answer - D | |
| 14. |
Calculate the grams of copper sulphate `(CuSO_(4))` needed to prepare `250.0 mL` of `1.00M CuSO_(4)` ?. |
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Answer» Moles of `CuSO_(4) = M xx V = 1 xx (250)/(1000)` Molar mass of copper sulphate = 159.6 g/mol Hence Mass of copper sulphate (gm) = Moles of `CuSO_(4) xx` Molar mass of copper sulphate `= 1 xx (250)/(1000) xx 159.6 g//mol` = 39.9 gm of Copper sulphate |
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| 15. |
A `100g NaOH` solution has `20g NaOH`. Find molality . |
| Answer» `m = (20//40)/(100 - 20) xx 1000 = (500)/(80) = 6.25 mol//kg` | |
| 16. |
What will be the Molarity of solution when water is added to `10g CaCO_(3)` to make `100mL` of solution ? . |
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Answer» Mol of `CaCO_(3) = 10 //100 = 0.1` Molarity = Mole of solute/Volume of solution (L) = 0.10 mol/0.10 L Therefore, Molarity of given solution = 1.0 M |
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| 17. |
Which of the following is correct match?A. A-i-IIB. B-iv-IC. C-iii-ID. D-ii-I |
| Answer» Correct Answer - B | |
| 18. |
34 g of hydrogen peroxide is present in 1135 mL of solution. Volume strength of solution isA. 10VB. 20VC. 30VD. 32V |
| Answer» Correct Answer - A | |
| 19. |
The volume strength of `1M H_(2)O_(2)` is (Molar mass of `H_(2)O_(2) = 34 g mol^(-1)`)A. 16.8B. 11.35C. 22.4D. 5.6 |
| Answer» Correct Answer - B | |
| 20. |
Statement-1 : Molarity of a solution depends on temperature but molality is independent of temperature Statement-2 : Molarity depends on volume of solution but molality depends on mass of solventA. Statement-1 is true, statement -2 is true and statement-2 is correct explanation for statement-3B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-3C. Statement-1 is false, statement-2 is trueD. Statement-1 is true, statement-2 is false |
| Answer» Correct Answer - A | |
| 21. |
For preparing 0.1 M solution of `H_(2)SO_(4)` in one litre, we need `H_(2)SO_(4)`A. `0.98g`B. `4.9g`C. `49.0g`D. `9.8g` |
| Answer» Correct Answer - D | |
| 22. |
1000 gram aqueous solution of `CaCO_(3)` contains 10 gram of carbonate. Concentration of solution is:A. 10 ppmB. 100 ppmC. 1000 ppmD. 10,000 ppm |
| Answer» Correct Answer - D | |
| 23. |
Calculate the mole fractions of the components of the solution composed by `92g` glycerol and `90g` water ? (M(water) = 18 , M (glycerol) = 92) . |
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Answer» Moles of water = 90 g /18 g = 5 mol water Moles of glycerol = 92g/92g = 1 mol glycerol Total moles in solution `= 5 + 1 = 6` mol Mole fraction of water = 5 mol/6 mol = 0.833 Mole fraction of glycerol = 1 mol/6mol = 0.167 |
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| 24. |
A solution is made by mixing `300ml 1.5M Al_(2) (SO_(4))_(3) + 300ml 2M CaSO_(4) + 400ml 3.5M CaCl_(2)` Find final molarity of `(1) SO_(4).^(-2), (2) Ca^(2+)` (3) `Cl^(-)` [Assume complete dissociation of these compound] . |
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Answer» (i) `[SO_(4)^(-2)]_(f) = ("Total moles")/("Total volume") = (300 xx 1.5 xx 10^(-3) xx 3 + 300 xx 2 xx 10^(-3))/((300 + 300 + 400)xx 10^(-3)) = 1.95 M` (2) `[Ca^(+2)]_(f) = (300 xx 2 + 400 xx 3.5)/(1000) = 2M` (3) `[Cl^(-)]_(f) = (400 xx 3.5 xx 2)/(1000) = 2.8 M` |
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| 25. |
Assuming complete precipitation of `AgCl`, calculate the sum of the molar concentration of all the ions if 2 lit of 2 M `Ag_(2)SO_(4)` is mixed with 4 lit of 1 M NaCl solution isA. 4 MB. 2 MC. 3 MD. 2.5 M |
| Answer» Correct Answer - B | |
| 26. |
Which of the following is correct match ?A. A-ii-IIIB. B-ii-IIIC. C-ii-IIID. D-ii-III |
| Answer» Correct Answer - D | |
| 27. |
`6.02 xx 10^(21)` molecules of urea are present in 100ml of its solution. The concentration of urea solution isA. `0.001M`B. `0.01M`C. `0.02M`D. `0.01M` |
| Answer» Correct Answer - D | |
| 28. |
`6.02 xx 10^(21)` molecules of urea are present in 100ml of its solution. The concentration of urea solution isA. `0.001 M`B. `0.01 M`C. `0.02 M`D. `0.1 M` |
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Answer» Correct Answer - D Moles of urea `= (6.02 xx 10^(21))/(6.02 xx 10^(23)) = (1)/(100)` Molarity `= (1000)/(100 xx 100) = 0.1 M` |
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| 29. |
A `5.2` molal aqueous of methyl alcohol, `CH_(3)OH`, is supplied. What is the molefraction of methyl alcohol in the solution ?A. `0.088`B. `0.050`C. `0.100`D. `0.190` |
| Answer» Correct Answer - A | |
| 30. |
A `5.2` molal aqueous of methyl alcohol, `CH_(3)OH`, is supplied. What is the molefraction of methyl alcohol in the solution ?A. `0.086`B. `0.050`C. `0.100`D. `0.190` |
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Answer» Correct Answer - A `5.2 m CH_(3)OH` ie 5.2 mol `CH_(3)OH` present in 1 kg water `X_(M) = (5.2)/(5.2 + (1000)/(18)) = (5.2)/(5.2 + 55.5) = (5.2)/(60.7) = 086` |
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| 31. |
Find the `% w//v " of " '10V' H_(2)O_(2)` solution |
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Answer» Molarity (M) of solution `= ("volume strength")/(11.35) = (10)/(11.35)` `% ((w)/(v)) = (M xx mol. Wt. " of solute")/(10) = (10)/(11.35) xx (34)/(10) = 3%` |
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| 32. |
Find the % labelling of 100 gm oleum sample if it contains 20 gm `SO_(3)` |
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Answer» % labelling of oleum sample `= (100 + x)%` `{:(SO_(3) " + " H_(2)O" "rarr,H_(2)SO_(4)),(20gm,),(1//4 " mole " 1//4 " mole ",),(" "4.5 gm,):}` `:.` % labelling of oleum sample `= (100 + 4.5) % = 104.5%` |
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| 33. |
Find the mass of Free `SO_(3)` present in 100gm, 109% oleum sample |
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Answer» 109% means, 9gm of `H_(2)O` is required `{:(SO_(3),+,H_(2)O,rarr,H_(2)SO_(4)),(,,9gm,,),(1//2 "mole",,1//2 "mole",,),(40gm,,,,):}` `:.` Mass of free `SO_(3) = 40 gm`, Mass of `H_(2)SO_(4) = 60 gm` Note: Work out, what are the maximum and minimum value of the % labelling |
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| 34. |
31 gm `H_(2)SO_(4)` is mixed with 20 gram `SO_(3)` to form mixture Determine % labelling of oleum solutionA. `104.5`B. 106C. 109D. 110 |
| Answer» Correct Answer - C | |
| 35. |
If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option isA. The resulting solution contains 18 gm of water and 118 gm `H_(2)SO_(4)`B. The resulting solution contains 9 gm water and 59 gm `H_(2)SO_(4)`C. The resulting solution contains only 118 gm pure `H_(2)SO_(4)`D. The resulting solution contains 68 gm of pure `H_(2)SO_(4)` |
| Answer» Correct Answer - B | |
| 36. |
`H_(2)O_(2)` solution used for hair bleacging is sold as a solution of approximately 5.0 g `H_(2)O_(2)` per 100 mL of the solution. The molecular mass of `H_(2)O_(2)` is 34. The molarity of this solution is approximatelyA. `0.15 M`B. `1.5 M`C. `3.0 M`D. `3.4 M` |
| Answer» Correct Answer - B | |
| 37. |
If 18 g of glucose is present in 1000 g of solvent, the solution is said to beA. 1 molarB. 0.1 molarC. 0.5 molarD. 0.1 molal |
| Answer» Correct Answer - D | |
| 38. |
171 g of cane sugar `(C_(12) H_(22) O_(11))` is dissolved in 1 litre of water. The molarity of the solution isA. `2.0 M`B. `1.0 M`C. `0.5 M`D. `0.25 M` |
| Answer» Correct Answer - C | |
| 39. |
8g NaOH is dissolved in one litre of solution, its molarity isA. 0.8 MB. 0.4 MC. 0.2 MD. 0.1 M |
| Answer» Correct Answer - C | |
| 40. |
Equal weight of `NaCl` and `KCl` are dissolved separately in equal of solutions. Molarity of the two solutions will be:A. EqualB. Greater for NaClC. Greater for KClD. Uncomparable |
| Answer» Correct Answer - B | |