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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. |
`K_(f)` for water is `1.86 K kg mol^(-1)`. IF your automobile radiator holds `1.0 kg` of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?A. 72 gB. 93 gC. 39 gD. 27 g |
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Answer» Correct Answer - B Coolant glycol `(C_(2)H_(6)O_(2))` is a non-electrolyte. `DeltaT_(1)=2.8^(@)` `DeltaT_(1)=(1000K_(f)w_(1))/(m_(1)w_(2))rArr2.8=(1000xx1.86xxw_(1))/(62xx1000)` `:.w_(1)=93.33g` |
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| 2. |
`20`g of a binary electrolyte(mol.wt.=`100`)are dissolved in `500`g of water.The freezing point of the solution is `-0.74^(@)CK_(f)=1.86K "molality"^(-1)`.the degree of ionization of the electrolyte isA. `0%`B. `100%`C. `75%`D. `50%` |
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Answer» Correct Answer - A `DeltaT=(1000xxK_(f)xxw)/(mxx500)` `0.74=(1000xx1.86xx20)/(mxx500)` `mxx100` `"Actual molecular mass"=100rArri=(100)/(100)=1+alpha` `alpha=0` `:.` The degree of ionisation of the electrolyte is 0%. |
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| 3. |
When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` isA. `75%`B. `50%`C. `25%`D. `85%` |
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Answer» Correct Answer - A |
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| 4. |
Two solutions of HCl, A and B have concentration of 0.5 N and 0.1 M respectively . The volume of solutions A and B required to make 2 L of 0.2 N HCl areA. `0.5L" of " A+1.5L" of " B`B. `1.5L" of " A+0.5L" of " B`C. `1.0L" of " A+1.0L" of " B`D. `0.75L" of " A+1.25L" of " B` |
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Answer» Correct Answer - A `N_(A)=0.5,V_(A)=VL("let")` `N_(B)=M_(B)=0.1,V_(B)=(2-V)L` From normality equation, `NV=N_(A)V_(A)+B_(B)V_(B)` `0.2xx2=0.5xxV+0.1xx(2-V)` `0.4=0.5V+0.2-0.1V` `0.4V=0.2` `V=(0.2)/(0.4)=0.5` `:.V_(A)=0.5L` `" and "V_(B)=(2-05)=1.5L` |
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| 5. |
Which of the following concentration terms is/are independent of temperature ?A. MolalityB. Molality and mole fractionC. Molality and mole fractionD. Molality and normality |
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Answer» Correct Answer - B Molality and mole fraction depends only upon weight but not on volume. Hence, these are not affected by change in temperature. |
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| 6. |
What Is the number of moles of `H_(2)SO_(4)` required to prepare `5.0L" of a"2.0M` solution of `H_(2)SO_(4)` ?A. `10`B. `5.0`C. `20`D. `2.5` |
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Answer» Correct Answer - A `"Number of moles = Molarity x Volume (in L)"` `rArr"Number of moles of "H_(2)SO_(4)=2.0xx5.0=10" moles"` |
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| 7. |
20 mL of HCI soution requires 19.85 mL of 0.01 M NaOH solution for complete neutralization . The molarity of HCI solution is _______M.A. `0.0099`B. `0.0099`C. `0.99`D. `0.1M` |
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Answer» Correct Answer - A `M_(1)V_(1)=M_(2)V_(2)` `20xxM_(HCI)=19.85xx0.01` `M_(HCI)=(19.85xx0.01)/(20)=0.0099` |
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| 8. |
Lowering of vapour pressure is highest forA. `0.1MBaCl_(2)`B. `0.1M" glucose"`C. `0.1MMgSO_(4)`D. urea |
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Answer» Correct Answer - A `(p^(@)-p_(s))/(p^(@))="molality"xx(1-alpha+xalpha+yalpha)` The value of `p^(@)-p_(s)` is maximum for `BaCl_(2)` . |
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| 9. |
Osmotic pressure present in the fluid inside the blood cell is equivalent toA. `0.9%(m//V)NaCl" solution"`B. `"less than "0.9%(m//V)NaCl" solution"`C. `"more than "0.9%(m//V)NaCl" solution"`D. `0.9%(m//V)Na_(2)SO_(4)" solution"` |
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Answer» Correct Answer - A Osmotic pressure associated with the fluid inside the blood cell is equivalent to that of `0.9%` (mass/ volume ) sodium chloride solution called normal saline solution and it is safe to inject intravenously. |
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| 10. |
Which of the following is not a colligative property?A. Depression in freezing pointB. Elevation in boiling pointC. Optical activityD. Relative lawering in vapour pressure |
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Answer» Correct Answer - C Optical activity is not a colligative property because is does not depends upon the number of solute particle, it depends on the nature of solutes. |
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| 11. |
Mark the correct option for the `K_(f)` . I. `K_(f) `depends on nature of solvent. II. `K_(f)` is also known as freezing point depression constant or molal depression constant. III. `K_(f)` is known as cryoscopic constant. IV. `"Unit of "K_(f)=Kkg" mol"^(-1)`. V. `M_(2)=(K_(f)xxW_(1)xx1000)/(DeltaT_(f)xxW_(2))`. where`W-(2)` is the mass of solute haveing molar mass of `M_(2)` present in `W_(1)` gram of solvent.A. I, II and IIIB. I, II, III and IVC. I. II, III, IV and VD. II, III, IV and V |
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Answer» Correct Answer - B (V) option is wrong correct formula : `M_(2)=(K_(f)xxw_(2)xx1000)/(DeltaT_(f)xxw_(1))` |
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| 12. |
Colligative properties of a solution depends uponA. nature of soluteB. nature of both solute and solventC. number of solute particlesD. number of solvent particles |
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Answer» Correct Answer - C The properties of solution which depend only on the number of solute particles but not on the nature of the solute taken are known as colligative properties. |
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| 13. |
To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` isA. 2 gB. `0.5g`C. 1 gD. `0.75g` |
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Answer» Correct Answer - C Elevation of boiling point, `DeltaT_(b)=(wxxK_(b)xx1000)/(MxxW)` `0.05=(wxx0.5xx1000)/(100xx100)orw=1g` |
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| 14. |
Mole fraction of vapour of `A` above solution in mixture of `A` and `B(X_(A) = 0.4)` will be `(P_(A)^(@) = 100mm, P_(B)^(@) = 200 mm)`:A. 0.4B. 0.25C. 0.85D. None of these |
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Answer» Correct Answer - B Mole fraction of A in vapour phase `=(p_(A)^(@)x_(A))/(p_("Total"))=(100xx10^(0.4))/(100xx4+200xx0.6)=0.25` |
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| 15. |
The density of `3 M` sodium of thiosulphate solution `(Na_(2)S_(2)O_(3))` is `1.25 g mL^(-1)`. Calculate a. The precentage by weight of sodium thiosulphate. b. The mole fraction of sodium thiosulphate. c. The molalities of `Na^(o+)` and `S_(2)O_(3)^(2-)` ions.A. `12.65%`B. `37.92%`C. `0.87%`D. `63.21%` |
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Answer» Correct Answer - B `"Molecular weight of "Na_(2)S_(2)O_(3)=158g" mol"^(-1)` From, molarity `=(%"by weight of solute" xx"density of solution"xx10)/(M)` `3=(%"by weight of"Na_(2)S_(2)O_(3)xx1.25xx10)/(158)` `:.%"by weight of "Na_(2)S_(2)O_(2)=(3xx158)/(1.25xx10)=37.92%` |
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| 16. |
The vapour pressure of water at `20^(@)C is 17.54mm`. When 20g of non - ionic substance is dissolved in 100g of water, the vapour pressure is lowered by `0.30mm`. What is the molecular mass of the substance ?A. `200.8`B. `206.88`C. `210.5`D. `215.2` |
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Answer» Correct Answer - B Relative lowering in vapour pressure, `(p^(@)-p_s)/(p^(@))=(x_(2))/(x_(2)+x_(1))=(w//m)/(w/m+W/M)` `(0.30mm)/(17.54mm)=(20//m)/(20//m+100//18)` `rArrm=206.88` |
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| 17. |
The following statements is true only under some specific conditions. Write the condition for it "Two volatile and miscible liquids can be separated by fractional distillation into pure components." |
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Answer» Correct Answer - T |
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| 18. |
Osmotic pressure of insulin solution at 298 K is found to be `0.0072atm`. Hence, height of water Column due to this pressure isA. `0.76cm`B. `0.70cm`C. `7.4cm`D. `76cm` |
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Answer» Correct Answer - C `p=0.0072atm=0.0072xx76"cm of Hg"` `pi=hdg,pi=0.0072xx76xx13.6cm" of mercury column Also,"p=hxx1cm" of water column"` `hxx1=0.0072xx76xx13.6cm` `:.h=0.0072xx76xx13.6cm=7.4cm` |
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| 19. |
60g of urea is dissolved in 1100g solution. To keep `DeltaT//K_(f)" as "1" mol"//kg`, water separated in the form of ice isA. 40 gB. 60 gC. 100 gD. 200 g |
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Answer» Correct Answer - A 1100g" solution has 60 g urea. Thus", `"water=1040g` `DeltaT_(f)=(1000K_(f)w_(1))/(m_(1)w_(2))rArr(DeltaT_(f))/(K_(f))=(1000xx60)/(60xxw_(2))=1` `:.w_(2)=1000g` `"Thus, ice formed"=1040-1000=40g` |
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| 20. |
At `10^(@)C`, the osmotic pressure of urea solution is `500 mm`.The solution is diluted and the temperature is raised to `25^(@)C`.when the osmotic pressure is found to be `105.3 mm`. Determine the extent of dilution.A. `V_("final")=5V_("initial")`B. `V_("initial")gtV_("final")`C. `V_("final")=4V_("initial")`D. `V_("final")=6V_("initial")` |
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Answer» Correct Answer - A For initial solution, `becausepi=(500)/(760)atm,T=283K` `:.(500)/(760)xxV_("Initial")=nxxRxx283` . . . (i) After dilution `:. (105)/(760)xxV_("Final")=nxxRxx298` . . . (ii) From Eqs. (i) and (ii), we get `(V_("Initial"))/(V_("Final"))=(1)/(5)i.e.` solution was diluted to 5 times. |
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| 21. |
What happens to freezing point of benzene when nephthalens is added ?A. IncreasesB. DecreasesC. Remains unchangedD. First decreases and then increases |
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Answer» Correct Answer - B Freezing point of a pure solvent decreases on addition of solute. |
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| 22. |
`1xx10^(3)m" solution of Pt "(NH_(3))_(4)Cl_(4) " in "H_(2)O` shows depression in freezing point by `0.0054^(@)C`. The structure of the compound will be `("given "K_(f)(H_(2)O)=1.860km^(-1))`A. `[Pt(NH_(3))_(4)]Cl_(4)`B. `[Pt(NH_(3))_(3)Cl]Cl_(3)`C. `[Pt(NH_(3))_(2)Cl_(2)]Cl_(2)`D. `[Pt(NH_(3))Cl_(3)]Cl` |
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Answer» Correct Answer - C `DeltaT_(f)=iK_(f)m` `0.0054=ixx1.86xx0.001` `i=(5.4)/(1.86)=2.90~=3` `So,[Pt(NH_(3))_(2)Cl_(2)]Cl_(2) " is the structure of the compound"` `[Pt(NH_(3))_(2)Cl_(2)]Cl_(2)hArr[Pt(NH_(3))_(2)Cl_(2)]^(o+)+2Cl^(Θ)` |
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| 23. |
At `40^(@)C` the vapour pressure of pure liquids, benzene and toluene, are `160 mm Hg` and `60 mm Hg` respectively. At the same temperature, the vapour pressure of an equimolar solution of the liquids, assuming the ideal solution will be:A. 140mm HgB. 110mm HgC. 220mm HgD. 100mm Hg |
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Answer» Correct Answer - B `p_("total")=p_(1)^(@)x_(1)+p_(2)^(@)x_(2)` `160xx(1)/(2)+60xx(1)/(2)=110mm" Hg"` |
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| 24. |
At 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg, if one more mole of B is added to the above ideal solution at 300K. Then the vapour pressure of A in its pure state isA. 300mm of HgB. 400mm of HgC. 500mm of HgD. 600mm of Hg |
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Answer» Correct Answer - A `p_(t)=p_(A)x_(A)+p_(B)x_(B)` `"In lst case, "500=(1)/(3)p_(A)+(2)/(3)p_(B)` `"or "1500=p_(A)+2p_(B)` . . . (i) in llnd case, (when one more mole of B isadded ) `2100=p_(A)+3p_(B)` From Eqs. (i) and (ii) `p_(B)=600mm" Hg and p_(A)=300mm Hg"` |
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| 25. |
A 6% of urea is isotonic withA. 1 M solution of glucoseB. `0.05M` solution of glucoseC. `6%` solution of glucoseD. `25%` solution of glucose |
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Answer» Correct Answer - A `"Molarity of urea"= ((6)/(60))/((100)/(1000))` Hence, 1m solution of glucose is isotonic with `6%` yrea solution. |
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| 26. |
How much `C_(2)H_(5)OH` must be added to `1.0"L of "H_(2)O`, so that solution should not freeze at `-4^(@)F` ? `[K_(f)(C_(2)H_(5)OH)=1.86^(@)C//m]`A. `lt10.75g`B. `gt494.5g`C. `lt20g`D. `494.5g` |
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Answer» Correct Answer - B `(C)/(5)=(F-32)/(9)` `-4^(@)F=-20^(@)CrArrDeltaT_(f)=20^(@)C` `"From "DeltaT_(f)=K_(f).mrArrm=(20)/(1.86)=10.75m` `"Mass of "C_(2)H_(5)OH" needed "=10.75xx46=494.5g` Solution not to freeze at `-20^(@)C` will require more ethanol than the mass needed to freeze at `-20^(@)C`. |
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| 27. |
A sample ofsea water contains `5xx10^(-3)g` of dissolved oxygen in 1 kg of the sample. The concentration of `O_(2)` in that sea water sample in ppm isA. `5xx10^(-4)`B. `5xx10^(-3)`C. 5D. `5xx10^(-1)` |
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Answer» Correct Answer - C `"Given",10^(3)g(1kg)"sample contains oxygen"=5xx10^(-3)g` `:.10^(6)g` sample will contain oxygen `=(5xx10^(-3)g)/(10^(3)g)xx10^(6)g=5"ppm"` |
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| 28. |
A mixture of ethane and ethene occupies 41 L at atm and 500 K. The mixture reacts compeletly with 10/3 mole of oxygen to produce `CO_(2)` and water. The mole fraction of ethane and ethene in the mixture are (R=0.0821L atm `K^(-1)mol^(-1)` respectivelyA. `0.50,0.50`B. `0.75,0.25`C. `0.67,0.33`D. `0.25,0.75` |
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Answer» Correct Answer - C `pV=nRT` `1xx41=nxx0.0821xx500rArrn=0.998mol` The number of moles of ethane ` = x` So number of moles of ethene `=(0.998-x)` Reaction of ethane and ethene with `O_(2)` (i) `2C_(2)H_(6)+70_(2)to4CO_(2)+6H_(2)O` (ii) `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O` x mole of ethane reacts `"with"(7x)/(2)"moles of " O_(2)` `(0.998-x)` mole of ethene reacts `=3(0.998-x)"mole of "O_(2)` `(7x)/(2)+[3(0.998-x)]=(10)/(3)"mole of "O_(2)` `x=(0.3393)/(0.5)=0.678"mole of ethane"` `"mole of ethene"=0.998-0.678=0.32` `x_("enthane")=(eta_("ethane"))/(eta_("ethane")+eta_("ethene"))=(0.678)/(0.678+0.32)=0.67` `x_("ethene")=1-x_("ethane")=1-0.67=0.33` |
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| 29. |
Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)A. 94.39mmB. 141.93mmC. 134.44mmD. 199.34mm |
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Answer» Correct Answer - B Relative lowering in vapour pressure, `(p^(@)-p_(s))/(p_(@))=(wxxM)/(mxxW)` `(143-p_(s))/(143)=(0.5)/(65)xx(154)/(1.58xx100)" "[because"molecular weight of "C Cl_(4)=154 " and weight = density " xx" volume"]` `143-p_(s)=1.07rArrp_(s)=141.93mm` |
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| 30. |
By dissolving 5 g substance in 50 g of water, the decrease in freezing point is `1.2^(@)C` . The gram molal depression is `1.85^(@)C` . The molecular weight of substance isA. `105.4`B. `118.2`C. `137.2`D. `154.2` |
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Answer» Correct Answer - D We know that `m=(1000K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))` `:.m=(1000xx1.85xx5)/(1.2xx50)=154.2` |
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| 31. |
`1`mol each of the following solutes are taken in `5`mol water, (a)`NaCl` (b)`K_(2)SO_(4)`( C)`Na(3)PO_(4)`(d) glucose Assuming `100%`ionisation of the electrolyte ,relative decrease in vapour pressure will be in orderA. NaClB. `K_(2)SO_(4)`C. `Na_(3)PO_(4)`D. glucose |
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Answer» Correct Answer - C Relative decrease in vapour pressure `propto` mole fraction of solute. `{:("Solute","Mole in solution","Mole fraction of solute"),(A(NaCl)," "2," "(2)/(7)=0.29),(B(K_(2)SO_(4))," "3," "(3)/(8)=0.38),(C(N_(3)PO_(4))," "4," "(4)/(9)=0.44),(D("Glucose")," "1," "(1)/(6)=0.17):}` Hence, `DltAltBltC` |
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| 32. |
A solution of `Al_(2)(So_(4))_(3){d=1.253 gm//ml}` contain `22%` salt by weight. The molarity , normality and molality of the solution isA. `0.805M,0.825m`B. `0.825M,0.805m`C. `4.83M,4.83m`D. `4.83M,48.3m` |
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Answer» Correct Answer - A `"Molarity"=(%xx10xxd)/("gram molecular mass")` `=(22xx10xx1.253)/(324)=0.806M` `"Molality"=(22xx1000)/(342(100-22))=0.825m` |
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| 33. |
Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:A. `1.14"mol kg"^(-1)`B. `3.28"mol kg"^(-1)`C. `2.28"mol kg"^(-1)`D. `0.44"mol kg"^(-1)` |
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Answer» Correct Answer - C `"Molality "(m)=(M)/(1000 " "d-MM_(1))xx100` `M="Molarity"," "M_(1)="Molecular mass"," "d="density"` `(2.05)/((1000xx1.02)-(2.05xx60))=1000` `=2.28mol" "kg^(-1)` |
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| 34. |
The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute, is 0.0125. The molality of the solution isA. `0.70`B. `0.50`C. `0.90`D. `0.80` |
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Answer» Correct Answer - A `(p^(@)-p_(s))/(p^(@))=x_(2)rArr(p^(@)-p_(s))/(p^(@))=(wN)/(mW)` `0.0125=(wM)/(mW)rArr(w)/(mW)=(0.0125)/(18)=0.00069` `"Hence, molality"=(w)/(mW)xx1000=0.00069xx1000=0.69` `~=0.70` |
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| 35. |
The mass of a non-volatile solute of molar mass `40g" "mol^(-1)` that should be dissolved in 114 g of octane to lower its vapour pressure by 20% isA. `11.4g`B. `9.8g`C. `12.8g`D. 10g |
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Answer» Correct Answer - D `"Given, "p^(@)=100,p=100-20=80`, `M_(2)=40" gmol"^(-1),M_(1)("octane")=C_(8)H_(18)=114" gmol"^(-1)` `w_(2)=?,w_(1)("octane")=114g` `(p^(@)-p)/(p)=(w_(2))/(M_(2))xx(M_(1))/(w_(1)` `(100-80)/(80)=(w_(2))/(40)xx(114)/(114)` `rArrw_(2)=10g` |
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| 36. |
Benzene and naphthalene form an ideal solution at room temperature. For this process,the true statement(s) is (are)A. `DeltaG" is positive " `B. `DeltaS_("system")"is positive " `C. `DeltaS_("surroundings")=0`D. `DeltaH=0` |
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Answer» Correct Answer - D When an ideal solution is formed process is spontaneous, thus `DeltaGlt0,DeltaSgt0,DeltaH=0andDeltaS_("Surrounding")lt0` |
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| 37. |
What is the volume of ethyl alcohol (density 1.15 g/cc) that has to be added to prepare 100 cc of 0.5 M ethyl alcohol solution in water ?A. `1.15c c`B. `2c c`C. `2.15c c`D. `2.30c c` |
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Answer» Correct Answer - B `" Molarity "=("mass"xx1000)/(" molecular weight " xx" volume of solution ")` Molecular weight of ethyl alohol, `C_(2)H_(5)OH=46g" mol"^(-1)` `0.5M=("mass"xx1000)/(46xx100)` `rArr"mass"=2.3g` `:.` Volume of ethyl alcohol required, `V(" mass of ethyl alcohol ")/("demsity of ethyl alcohol")` `(2.3g)/(1.15g//c c)=2c c` |
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| 38. |
What will be the value of molality for an aqueous solution of 10% w/W NaOH?A. 5B. `2.778`C. 10D. `2.5` |
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Answer» Correct Answer - B `"Moles of "NaOH=(10)/(40)=(1)/(4)=0.25` `:."Molality "(m)=("Moles of "NaOHxx1000)/("Weight of water (in g)")=(0.25xx1000)/(90)` `=2.778` |
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| 39. |
What is the osmotic pressure of `12%` solution of can sugar (mol.wt. 342) at `17^(@)C`A. `2.42"atm"`B. `4.33"atm"`C. `8.35"atm"`D. `16.30"atm"` |
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Answer» Correct Answer - C 12 g of sugar is dissolved in 100 mL water. `:.` 342g of sugar is dissolved in `(100)/(12)xx342=2850mL=2.85L` `:.V=2.85LrArrpiV=RT` `"or "pi=(RT)/(V)=(0.082xx290)/(2.85)=8.343"atm"` `~=8.35"atm"` |
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| 40. |
An aqueous solution freezes at `-0.186^(@)C (K_(f)=1.86^(@)` ,`K_(b)=0.512^(@)`. What is the elevation in boiling point?A. `0.0512^(@)C`B. `100.0512^(@)C`C. `-0.0512^(@)C`D. None of these |
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Answer» Correct Answer - A Elevation in boilling point, `DeltaT_(b)=K_(b)xxm` Depression in freezing point `DeltaT_(f)=K_(f)xxm` `(DeltaT_(b))/(DeltaT_(f))=(K_(b))/(K_(f))and(DeltaT_(b))/(0.186)=(0.512)/(186)` `DeltaT_(b)=0.0512^(@)C` |
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| 41. |
How many grams of sucrose (molecular weight 342) should be dissolved in `100 g` water in order to produce a solution with `105^(@)C` difference between the freezing point and the boiling point ? `(K_(b) =0.51^(@)C m^(-1) , (K_(f) =1.86^(@)C m^(-1))`A. `34.2g`B. `72g`C. `342g`D. `460g` |
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Answer» Correct Answer - B `"Boiling point "(T_(b))=100+DeltaT_(b)=100+k_(b)m` `"Freezing point "(T_(f))=0-DeltaT_(f)=-k_(f)m` `T_(b)-T_(f)=(100+k_(b)m)-(-k_(f)m)` `105=100+0.51m+1.86m` `2.37m=5" or m"=(5)/(2.37)=2.11` `:."Weight of sucrose to be dissolved in 100g water"` `=(2.11xx342)/(1000)xx100=72g` |
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| 42. |
An antifreeze solution is prepared from `222.6 g` of ethylene glycol `[C_(2)H_(4)(OH)_(2)]` and `200 g` of water. Calculate the molality of the solution. If the density of the solution is `1.072g mL^(-1)` then what shall be the molarity of the solution?A. `9.10,17.95`B. `10.90,16.6`C. `12.04,17.95`D. `18.2,16.97` |
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Answer» Correct Answer - A `"Molality"=("moles of solute (glycol)")/("kg of solvent")=(222.6)/(62)/0.200` `=17.95" mol "kg^(-1)` `"Total mass of solution"=222.6+200=422.6g` `"Volume of solution"=("mass")/("density")=(422.6)/(1.072)mL=(0.4226)/(1.072)L` `:."Molarity"=("moles of solute")/("litre")=(22.6//62)/(0.4226//11.072)=9.10M` |
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| 43. |
Aluminium phosphate is `100%` ionised in `0.01` molal aqueous solution . Hence , `DeltaT_(b) // K_(b)` is :A. `0.01`B. `0.015`C. `0.0175`D. `0.02` |
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Answer» Correct Answer - D `AlPO_(4)hArrAl^(3)+PO_(4)^(3-)` `i=1+x=2` `DeltaT_(b)="molality"xxK_(b)xxi=0.01xxK_(b)xx2` `:.(DeltaT_(b))/(K_(b))=0.02` |
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| 44. |
The solubility of `N_(2)(g)`in water exposed to the atmosphere, when the partial pressure is `593`mm is `5.3xx10^(-4)`M. Its solubility at 760 mm and at the same temperature isA. `4.1xx10^(-4)M`B. `6.8xx10^(-4)M`C. `1500M`D. `2400M` |
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Answer» Correct Answer - B Solubility `propto" pressure",(S_(2))/(S_(1))=(p_(1))/(p_(1))` `S_(2)=5.3xx10^(-4)xx(760)/(593)` `=6.8xx10^(-4)M` |
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| 45. |
`5.5mg` of nitrogen gas dissolves in 180g of water at 273 K and 1 atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180g of water at 5 atm nitrogen pressure is approximatelyA. `1xx10^(-6)`B. `1xx10^(-5)`C. `1xx10^(-3)`D. `1xx10^(-4)` |
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Answer» Correct Answer - B Amount of nitrogen gas at 5 atm `=5xx5.5=27.5mg` `=27.5xx10^(-3)g` `"Mole fraction of "N_(2)=("Moles of "N_(2))/("Moles of "H_(2)O+"Moles of "N_(2))` `("Moles of "N_(2))/("Moles of "H_(2)O)" "(becauseN_(2)lt ltH_(2)O)` `=((27.5xx10^(-3))/(28))/(180/(18))=1xx10^(-5)` |
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| 46. |
Assertion Increasing temperature of a volatile liquid increases its vapour pressure. Reason Increase in temperature increases the average kinetic energy of molecules in the vapour phase hence they collide to the surface of liquid with greater force.A. Both assertion and reason are correct and reason is the correct explanation of the assertion,B. Both assertion and reason are correct but reason is not the correct explanation of assertion.C. Assertion is correct but reason is wrong.D. Assertion is wrong but reason is correct. |
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Answer» Correct Answer - B::C::D |
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| 47. |
Mole fraction of the solute in a 1 molal aqueous solution is :A. `1.7700`B. `1.7770`C. `0.0177`D. `0.0344` |
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Answer» Correct Answer - C `"Mole fraction of solute"=(1)/(56.56)=0.0177` |
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| 48. |
An aqueous solution of glucose is `10%` in strength ,The volume in which `1`g mole of it dissolved will beA. `9N`B. `0.3N`C. `3N`D. `1N` |
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Answer» Correct Answer - A Orthophosphoric acid `(H_(3)PO_(4))` is a tribasic acid. `.: "Normality" ="molarity"xx"basicity"` `:."Normality"=3Mxx3=9N` |
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| 49. |
An aqueous solution of glucose is `10%` in strength ,The volume in which `1`g mole of it dissolved will beA. `18L`B. `9L`C. `0.9L`D. `1.8L` |
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Answer» Correct Answer - D `10%` glucose solution = 10g `=(10)/(180)"mole in "100c c,i.e.0.1L.` Hence, 1 mole of glucose will be present in ltbr. `=(0.1xx180)/(10)=1.8L` |
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| 50. |
Mole fraction of solute in benzene is `0.2` then what is the value of molality of solute?A. `3.2`B. 2C. 4D. `3.6` |
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Answer» Correct Answer - A Let the solution be x molal, then x moles of solute are present in `1000g" of benzene"=(1000)/(78)=12.82` benzene `:."Mole fractionof solute"=(x)/(x+12.82)` `rArr0.=(x)/(x+12.82)` `x=(2.564)/(0.8)=3.2` |
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