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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A sample of pure lead weighing `2.07 g` is dissolved in nitric acid to give a solution of lead nitrate. This solution is treated with hydrochloric acid, chlorine gas and ammonium chloride. The result is a precipitate of `(NH_(4))_(2) PbCl_(6)`. What is the maximum weight of this product that could be obtained form the lead sample? |
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Answer» Eq. of `Pb=Eq. "of" Pb(NO_(3))_(2)` `=Eq."of"(NH_(4))_(2)PbCl_(6)` `(2.07)/(207//2)=(w)/(456//2)` `:. Wt. "of" (NH_(4))_(2)PbCl_(6), w= 4.56 g` |
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| 2. |
Persons are medically considered to have lead poisoning if they have a concentration greater than `10` micrograms of lead per decilitre of blood. What is the concentration in parts per billion? |
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Answer» `10^(-1)` litre blood has `10xx10^(-6) g Pb` `10^(9)` litre blood has `(10xx10^(-6)xx10^(9))/(10^(-1))=10^(5)ppb` |
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| 3. |
A bottle is labelled `122.5%` oleum. `22.7 mL` of `Ca(OH)_(2)` is unknown molarity are used to completely neutralise `1 g` oleum. Find the normality of `Ca(OH)_(2)`. |
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Answer» Correct Answer - 2 `(100+22.5)%` oleum means `222.5 g H_(2)SO_(4)` `H_(2)O+SO_(3)rarrH_(2)SO_(4)` `because 18 gH_(2)O "gives" =98g H_(2)SO_(4)` `therefore 22.5 g H_(2)O"gives" =(98xx22.5)/(18)=122.5gH_(2)SO_(4)` `therefore` Total `H_(2)SO_(4)=100+122.5=222.5g` 1 g oleum equiv `2.225gH_(2)SO_(4)=(2.225)/(49)xx1000` `45.41 meq.` `therefore "Meq.of" H_(2)SO_(4)=45.41` `therefore22.7xxN=45.41` `N=2` |
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| 4. |
`V_(1) mL` of `NaOH` of normality `X` and `V_(2) mL` of `Ba(OH)_(2)` of mormality `Y` are mixed together. The mixture is completely neutralised by `100 mL` of `0.1 N HCl`. If `V_(1)//V_(2)=1/4` and `X/Y=4`, what fraction of the acid is neutralised by `Ba(OH)_(2)`?A. `0.5`B. `0.25`C. `0.33`D. `0.67` |
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Answer» Correct Answer - A `"Meq.of"NaOH+Meq.of Ba(OH)_(2)=Meq.of HCl` `XV_(1)+YV_(2)=100xx0.1=10` `4Yxx(V_(2))/(4)+YV_(2)=10` `2YV_(2)=10` `:. V_(2)Y=5` `:. V_(1)X=5` `:.` Fraction of acid by `Ba(OH)_(2)` `=(5)/(10)=0.5` |
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| 5. |
What is the total molar concentration of ions in `0.350M` solution of `Na_(2)SO_(4)` assuming its complete isonisation? |
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Answer» `Na_(2)SO_(4(s))overset(H_(2)O)rarr2Na_((aq.))^(+)SO_(4(aq.))^(2-)` Total molar concentration `=[Na^(+)]+[SO_(4)^(2-)]` `=(2xx0.35)+0.35= 1.05M` |
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| 6. |
`100 mL` each of `0.5 N NaOH, N//5 HCl` and `N//10H_(2)SO_(4)` are mixed together. The resulting solution will be:A. AcidicB. NeutralC. AlkalineD. none of these |
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Answer» Correct Answer - C `"Meq.of" NaOH=100xx0.5=50` `Meq."of"HCl=(1)/(5)xx100=20` `Meq."of"H_(2)SO_(4)=(1)/(10)xx100=10` Total Meq. of acid`=20+10=30` Total.Meq. of `NaOH=50` Meq. of `NaOH"left"=50-30=20` Thus resulting solution will be alkaline. |
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| 7. |
The number of atomic weight scale is based on:A. `C^(12)`B. `O^(16)`C. `H^(1)`D. `C^(13)` |
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Answer» Correct Answer - A It is a fact. |
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| 8. |
Amount of oxygen in `32.2 g` of `Na_(2)SO_(4).10H_(2)O` is:A. `20.8 g`B. `26.71 g`C. `2.24 g`D. `2.08 g` |
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Answer» Correct Answer - B Mol.wt.of `Na_(2)SO_(4).10H_(2)O=270` `270 g Na_(2)SO_(4).10H_(2)O-=224 gO` |
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| 9. |
A sample supposed to be pure `CaCO_(3)` is used to standardise a solution of `HCl`. The substance really was a mixture of `MgCO_(3)` and `BaCO_(3)`, but the standardisation of `HCl` was accurate. Find the percentage of `BaCO_(3)` and `MgCO_(3)` in mixture. |
| Answer» Correct Answer - `BaCO_(3)=27.89%, MgCO_(3)=72.11%`; | |
| 10. |
In the reaction: `2Al_((s))+6HCl_((aq.))rarr2Al_((aq.))^(3+)+6Cl_((aq.))^(-)+3H_(2(g))`A. `6 "litre" HCl_((aq.))` is consumed for every `3 LH_(2(g))` producedB. `33.6 "litre" H_(2(g))` is produced regardless of temperature and pressure for every mole `Al` that reactC. `67.2 "litre" H_(2(g))` at `STP` is produced for every mole `Al` that reactsD. `11.2 "litre" H_(2(g))` at `STP` is produced for every mole `HCl_((aq.))` consumed |
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Answer» Correct Answer - D `2Al_((s))+6HCl_((aq.))rarr2Al_((aq.))^(3+)6Cl_((aq.))^(-)+3H_(2_((g)))` For each mole of `HCl` reacted, `0.5` mole of `H_(2)` gas is formed at `STP`. `1` mole of an ideal gas occupies `22.4` litre at `STP` Volume of `H_(2)` gas formed at `STP` per mole of `HCl` reacted is `22.4xx0.5` litre. |
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| 11. |
`0.9698 g` of an acid are present in `300 mL` of a solution. `10 mL` of this solution requires exactly `20 mL` of `0.05 N KOH` solution. If the `mol.wt.` of acid is `98`, calculate the number of neutralizable protons. |
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Answer» Correct Answer - 3 Meq.of acid in `10 mL` solution `= "Meq.of" KOH` `=20xx0.05` `:.` Meq.of acid in `300 mL` solution `=(20xx0.05xx300)/(10)=30` `:. (0.9698)/(98//n)xx1000=30` ( where `n` is no.of neutralizable proton) `therefore n= 3` |
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| 12. |
`20 mL` of `0.1M H_(3)BO_(3)` solution on complete netralisation requires ….. mL of `0.05M NaOH` solution:A. `20 mL`B. `40 mL`C. `120 mL`D. `80 mL` |
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Answer» Correct Answer - B `H_(3)BO_(3)` is monobasic acid `"Meq. of"H_(3)BO_(3)="Meq.of" NaOH` `20xx0.1=Vxx0.05` `V=40 mL` |
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| 13. |
`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg(OH)_(2)`?A. `0.2 Eq.`B. `0.05 Eq.`C. `0.02 Eq.`D. `0.01 Eq.` |
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Answer» Correct Answer - B `0.05` equivalent or mole of `NaHCO_(3)` requires `0.05` eq. of `Mg(OH)_(2)`. `2NaHCO_(3)+Mg(OH)_(2)rarrMgCO_(3)+Na_(2)CO_(3)+2H_(2)O` |
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| 14. |
`100 mL` of `0.1 M` solution of `H_(2)SO_(4)` is used to prepare `0.05N` solution of `H_(2)SO_(4)`. What is the volume of water added to prepare the desired solution:A. `300 mL`B. `400 mL`C. `100 mL`D. `200 mL` |
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Answer» Correct Answer - A `"Meq. of dil". H_(2)SO_(4)="Meq. of conc". H_(2)SO_(4)` `100xx0.1xx2=0.05xxV` `:. V_("solution")=400mL` `:. V_("water")=400-100=300mL` |
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| 15. |
Find the concentration of `1.6N` solution of `H_(2)O_(2)` in terms of volume. |
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Answer» Correct Answer - 9 `because` Equivalent of `H_(2)O_(2)` in 1 litre solution `=1.6` `therefore w_(H_(2)O_(2))` in 1 litre solution `=1.6xx(34)/(2)=27.2 g` `2H_(2)O_(2)rarr2H_(2)O+O_(2)` `therefore` Volume of `O_(2)` obtained by `1000 mL` `H_(2)O_(2)` solution`=(22400xx27.2)/(68)=8960 mL` `therefore` Volume strength of `H_(2)O_(2)=(8960)/(1000)=8.96approx9.0` |
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| 16. |
Which of the following statements are correct?A. The equivalent weight of `Ba_(3)(PO_(4)) is 100.1`B. The equivalent weight of `Na_(3)PO_(4) is 54.66`C. The equivalent weight of `H_(3)PO_(4) is 32.67`D. The equivalent weight of `Ca(OH)_(2) is 36.5` |
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Answer» Correct Answer - A::B::C Equivalent wt.of compound `E=("Molecular wt.of compound")/("Total charge on cationic/anionic part")` `:. E_(Ba_(3)(PO_(4))_(2))=(M_(Ba_(3)(PO_(4))_(2)))/(6)=(601)/(6)=100.1` ( `:.` Total charge on cations or anions`=6`) Also say for `Ba_(3)(PO_(4))_(2)` `E_(Ba_(3)(PO_(4))_(2))=E_(Ba^(2+))+E_(PO_(4)^(3-))` `=(at.wt."of" Ba)/(2)+("radical wt.of"PO_(4)^(3-))/(3)` `(137)/(2)+(9)/(3)=100.1` `E_(Na_(3)PO_(4))=(M_(Na_(3)PO_(4)))/(3)=(164)/(3)=54.66` ( `:.` Total charge on cations or anions `=3`) `E_("Acid")=("mol.wt".)/("Basicity")` `:. E_(H_(3)PO_(4))=(98)/(3)=32.67` `E_("base")=("Mol.wt.")/("Acidity")` `E_(Ca(OH)_(2))=(71)/(2)=35.5` |
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| 17. |
Number of atoms in `558.5 g Fe (at.wt.55.85)` is:A. Twice that in `60 g` carbonB. `6.023xx10^(22)`C. Half in `8g He`D. `558.5xx6.023xx10^(23)` |
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Answer» Correct Answer - A Mole of `Fe=(558.5)/(55.85)=10`, Moles of `C=(60)/(12)=5` |
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| 18. |
Which of the following with increase in temperature?A. MolalityB. Weight fraction of soluteC. Fraction of solute present in waterD. Mole fraction |
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Answer» Correct Answer - C The volume of water increases with temperature. |
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| 19. |
Dissolving `120g` of urea `(Mw = 60)` in `1000 g` of water gave a solution of density `1.15 g mL^(-1)`. The molarity of solution is:A. `1.78 M`B. `1.02 M`C. `2.05 M`D. `0.50 M` |
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Answer» Correct Answer - C Mass of solute =`120 g` Molecular mass of solute `=60 u` Mass of solvent `=1000 g` Density of solution `=1.15 g//mL` Total mass of solution `=1000+120=1120 g` `"Volume of solution"=("Mass")/("Density")=(1120)/(1.15)mL` `"Molarity"=(("Mass of solute")/("Molecular mass of solute"))/("Volume of solution")xx1000` `=(120//60)/(1120//1.15)xx1000` `=(2xx1000xx1.15)/(1120)=2.05 M` |
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| 20. |
`6.02xx10^(20)` molecules of urea are present in `100 mL` solution. The concentration of urea solution is:A. `0.1 M`B. `0.01M`C. `0.02 M`D. `0.001 M` |
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Answer» Correct Answer - B `M=("Moles of urea")/(V"in litre")=(6.02xx10^(20)xx1000)/(6.023xx10^(23)xx100)` `10^(-2)M` |
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| 21. |
One litre of `N_(2)` and `7//8` litre of `O_(2)` under identical conditions of `P` and `T` are mixed. The amount of gases present in mixutre show:A. `w_(N_(2)) = 3w_(O_(2))`B. `w_(N_(2))=8w_(O_(2))`C. `w_(N_(2)) = w_(O_(2))`D. `w_(N_(2)) = 16 w_(O_(2))` |
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Answer» Correct Answer - C At constant pressure and temperature `V prop n` For `N_(2), 1 prop (w_(1))/(28)` For `O_(2),(w_(2))/(32)` |
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| 22. |
Total number of atoms present in `1.0 cm^(3)` of solid urea (density `0.3 g//cm^(3))` at `25^(@)C` are:A. `3.01 xx 10^(21)`B. `2.41xx10^(22)`C. `3.01 xx 10^(22)`D. `2.41 xx 10^(23)` |
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Answer» Correct Answer - B `w_("urea")=1xx0.3 g`, Also `60 g` urea has `8N` atoms `(NH_(2)CONH_(2))` |
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| 23. |
The domestic water supply is treated by bleaching powder to remove unhygienic species in water and to make it safe for drinking water. However, this given rise to `Ca^(2+)` and `Cl^(-)` ion contamination in water. Both these ions are also injurious for health if a minimum concentration is crossed. The chloride ions are tested by a kit provided by many companies having `AgNO_(3)` solution that is added drop by drop to `23 mL` of water sample to which an indicator has been added. When sufficient silver nitrate is added to remove `Cl^(-1)` ions as `AgCl` solid, the solid turns orange. The colour change is noticed by addition of `AgNO_(3)` having molar concentration such that each drop `(0.05 mL)` of `AgNO_(3)` converts `12.5 mg` of `Cl^(-)` ions `AgCl`. The molar concentration of `Cl^(-)` in the sample of water used is:A. `3.225xx10^(-3)M`B. `2.225 M`C. `1.521 M`D. `0.1837 M` |
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Answer» Correct Answer - D Thus, molarity of `Cl^(-)` in sample `=(6.52)/(35.5)=0.1837M` |
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| 24. |
The domestic water supply is treated by bleaching powder to remove unhygienic species in water and to make it safe for drinking water. However, this given rise to `Ca^(2+)` and `Cl^(-)` ion contamination in water. Both these ions are also injurious for health if a minimum concentration is crossed. The chloride ions are tested by a kit provided by many companies having `AgNO_(3)` solution that is added drop by drop to `23 mL` of water sample to which an indicator has been added. When sufficient silver nitrate is added to remove `Cl^(-1)` ions as `AgCl` solid, the solid turns orange. The colour change is noticed by addition of `AgNO_(3)` having molar concentration such that each drop `(0.05 mL)` of `AgNO_(3)` converts `12.5 mg` of `Cl^(-)` ions `AgCl`. If `12` drops of `AgNO_(3)` solution are used to reach the colour point, what mass of chloride ion is present in one litre sample?A. `6.52 g`B. `7.150 g`C. `5.125 g`D. `1.25g` |
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Answer» Correct Answer - A `1` drop `AgNO_(3)= 12.5xx10^(-3)g Cl^(-)` `therefore 12 "drop" AgNO_(3)=12.5xx10^(-3)xx12 "mole" Cl^(-)` `=150xx10^(-3)g Cl^(-1)` Also, `23 mL` sample `=0.15 gCl^(-1)` `therefore 1000 mL` sample `=(0.15xx1000)/(23)gCl^(-)//"litre solution"` |
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| 25. |
The domestic water supply is treated by bleaching powder to remove unhygienic species in water and to make it safe for drinking water. However, this given rise to `Ca^(2+)` and `Cl^(-)` ion contamination in water. Both these ions are also injurious for health if a minimum concentration is crossed. The chloride ions are tested by a kit provided by many companies having `AgNO_(3)` solution that is added drop by drop to `23 mL` of water sample to which an indicator has been added. When sufficient silver nitrate is added to remove `Cl^(-1)` ions as `AgCl` solid, the solid turns orange. The colour change is noticed by addition of `AgNO_(3)` having molar concentration such that each drop `(0.05 mL)` of `AgNO_(3)` converts `12.5 mg` of `Cl^(-)` ions `AgCl`. Assuming that concentration of `Ca^(2+)` ions in solution is equal equivalence ratio to chloride ions, the hardness of water is:A. `9.185xx10^(3)"ppm"`B. `6.185xx10^(3)"ppm"`C. `1.185xx10^(3)"ppm"`D. `4.185xx10^(3)"ppm"` |
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Answer» Correct Answer - A Eq. of `CaCO_(3)="Eq.of" Ca^(2+)= "Eq.of" Cl^(-)` ( for `1 "litre"` water) `(w)/(100//2)=0.1837` `w_(CaCO_(3))=(0.1837xx100)/(2)=9.185 g` `therefore` Hardness =`(9.185xx10^(6))/(10^(3))` `9.185xx10^(3)"ppm"` |
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| 26. |
The domestic water supply is treated by bleaching powder to remove unhygienic species in water and to make it safe for drinking water. However, this given rise to `Ca^(2+)` and `Cl^(-)` ion contamination in water. Both these ions are also injurious for health if a minimum concentration is crossed. The chloride ions are tested by a kit provided by many companies having `AgNO_(3)` solution that is added drop by drop to `23 mL` of water sample to which an indicator has been added. When sufficient silver nitrate is added to remove `Cl^(-1)` ions as `AgCl` solid, the solid turns orange. The colour change is noticed by addition of `AgNO_(3)` having molar concentration such that each drop `(0.05 mL)` of `AgNO_(3)` converts `12.5 mg` of `Cl^(-)` ions `AgCl`. The molar concentration of `AgNO_(3)` solution if one drop of `AgNO_(3)` measure `0.05 mL` is:A. `6.04xx10^(-3)M`B. `7.04xx10^(-3)M`C. `150 M`D. `3.52 M` |
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Answer» Correct Answer - B `1 "drop" AgNO_(3)= 0.05 mL AgNO_(3)` `therefore 12 "drop" AgNO_(3)=0.05xx12 mL AgNO_(3)` `= 0.60 mL AgNO_(3)` `mM "of" AgNO_(3)` in `12 "drop" = mM "of" Cl^(-)` in `23 mL` sample `Mxx0.6=(150xx10^(-3))/(35.5)` `therefore M=7.04xx10^(-3)` |
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| 27. |
`1.7 g` of silver nitrate dissolved in `100 g` of water is taken. `0.585 g` of sodium chloride dissolved in `100 g` of water is added to it an chemical reaction occurs. `1.435 g` of silver chloride and `0.85 g` of sodium nitrate are formed. Justify that the data obey law of conservation of mass. |
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Answer» Total mass before the cemical change `{:(=mass ofAgNO_(3),+,mass of NaCI,+,"mass of water"),(=1.7,+,0.585,+,200):}` `=202. 285 g` Total mass after the chemical reaction `{:(=mass ofAgCI,+,mass of NaNO_(3),+,"mass of water"),(=1.435,+,0.85,+,200):}` `=202.285 g` Thus total mass before reaction = total mass agter reaction. This confirms the law of conservation of mass. |
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| 28. |
How many molecule are present in one `mL` of water vapour of `STP` ? |
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Answer» `because 22.4` litre water vapour at `STP` has `= 6.023 xx 10^(23)` molecules `:. 1 xx 10^(-3)` litre water vapours at `STP` has `= (0.693 xx 10^(23) xx 10^(-3))/(22.4)` `= 2.69 xx 10^(19)` molecules |
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| 29. |
A plant virus is found to consist of uniform cylindrical particle of `150 Å` in diameter 5000 Å long. The specific volume of the virus is 0.75 `mLg^(-1)`. If the virus is considered to be a single particle, find its molar mass. |
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Answer» Volume of virus `= pir^(2)l` `=(22)/(7)=(150)/(2)xx(150)/(2)xx10^(-16)xx5000xx10^(-8)` `=0.884xx10^(-16)cm^(3)` `:.` Weight of one virus `=(0.884xx10^(-16))/(0.75)g` `=1.178xx10^(-16)g` `:.` Mol.wt.of virus `=1.178xx10^(-16)xx6.023xx10^(23)` `=7.095xx10^(7)` |
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| 30. |
How much `CO` is produced by the reaction of `1.0 kg` octane and `1.0 kg` oxygen. Also report the limiting reagent for this reaction. |
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Answer» `{:(,C_(8)H_(18)+,(17)/(2)O_(2)rarr,8CO+,9H_(2)O),("Moles before reaction",(10^(3))/(114),(10^(3))/(32),,),("Moles after reaction",[10^(3)/(114)-(10^(3)xx2)/(32xx17)],0,(8xx10^(3)xx2)/(32xx17),):}` Mole ratio for reaction is `C_(8)H_(18):O_(2)::1:(17)/(2)` and `O_(2):CO::(17)/(2):8` Thus, `O_(2)` is limiting reagent and moles of `CO` formed `= 29.41` Mass of `CO` formed `= 29.41xx28= 823.48 g` |
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| 31. |
Chlorophyll, the green colouring matter of plants responsible for photosynthesis, contains `2.68%` of magnesium by mass. Calculate the number of magnesium atoms in `2.00 g` of chlorophyll. |
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Answer» `:. 100 g` chlorophyll contains `2.68 g Mg` `=(2.68)/(64) "mole" Mg` `2 g` chlorophyll contains `(2.68xx2)/(24xx100)` `=2.2xx10^(-3) "mole"Mg` `:.` No. of `Mg "atoms" = 2.2xx10^(-3)xx6.023xx10^(23)` `=1.345xx10^(21)` atoms of `Mg` |
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| 32. |
A compound contains `10^(-2)%` of phosphorus. If atomic mass of phosphorus is `31`, the molar mass of the compound having one phosphorus atom per molecule is:A. `31`B. `31 xx 10^(2)`C. `31 xx 10^(4)`D. `31 xx 10^(3)` |
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Answer» Correct Answer - C `10^(-2) P` then mol.wt.of compound `= 100` `31 g P` then mol.wt.of compound `=(100xx31)/(10^(-2))=31xx10^(4)` |
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| 33. |
A drop `(0.05 mL)` of `12 M HCl` is spread over a thin sheet of aluminium foil ( thickness `0.10mm` and density of `Al=2.70g//mL`). Assuming whole of `HCl` is used to dissolve `Al`, what will be maximum area of hole produced in foil ? |
| Answer» Correct Answer - `0.2 cm^(2)`; | |
| 34. |
A hydrated sulphate of metal contained `8.1%` metal and `43.2%SO_(4)^(2-)` by weight. The specific heat of metal is `0.24 cal //g`. What is hydrated sulphate? |
| Answer» Correct Answer - `Al_(2)(SO_(4))_(3).18H_(2)O`; | |
| 35. |
The specific heat of a metal is `0.836 J//g`. The approximate at.wt.is:A. `16`B. `64`C. `40`D. `32` |
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Answer» Correct Answer - D `"At.wt."xx"specific heat" ("in" cal//g)=6.4` |
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| 36. |
How many `g`-atom are in `84 g` of carbon? |
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Answer» Correct Answer - 7 `because g-"atom"=("wt".)/("at.wt".)=(84)/(12)=7` |
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| 37. |
How many `g`-atom of `S` are present in `196 g` of `H_(2)SO_(4)?` |
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Answer» Correct Answer - 2 `because98 g H_(2)SO_(4)equiv32 g S=1 g-"atom of"S` `therefore 196 g H_(2)SO_(4)=(1xx196)/(98)=2 g-"atoms of"S` |
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| 38. |
What weight of `Na_(2)CO_(3) of 95%` purity would be required to neutralize `45.6 mL` of `0.235 N` acid? |
| Answer» Correct Answer - `0.5978g`; | |
| 39. |
How much water is to be added to dilute `10 mL` of `10 N HCl` to make it decinormal?A. `990 mL`B. `1010 mL`C. `100 mL`D. `1000 mL` |
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Answer» Correct Answer - A Meq. of conc.`HCl= "Meq.of dil".HCl`, `10xx10=Vxx(1)/(10)` `V=1000mL` Thus `990 mL` of water should be added to `10 mL` on conc. `HCl` to get decinormal solution. |
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| 40. |
`25 mL` of a solution of barium hydroxide on titration with `0.1 "molar"` solution of hydrochloric acid give a titre value of `35 mL`. The molarity of barium hydroxide is:A. `0.28`B. `0.35`C. `0.07`D. `0.14` |
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Answer» Correct Answer - C Meq.of `Ba(OH)_(2)="Meq.of" HCl` `Nxx25=0.1xx35` `therefore N_(Ba(OH)_(2))=(3.5)/(25)` `therefore M_(Ba(OH)_(2))=(3.5)/(25)xx(1)/(2)=0.07` |
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| 41. |
Commercially availiable concentrated hydrochloric acid contains `38% HCl` by mass. (a) What is the molarity of this solution? The density is `1.19g mL^(-1)` ? (b) What volume of concentrated `HCl` is required to make `1.00 "litre"` of `0.10M HCl`? |
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Answer» (a) `"Mass of" HCl= 38 g`, `"density of solution"=1.19g//mL`, `"Mass of solution" -100 g` `"Molarity"=("Moles of"HCl)/("Volume of solution in litre")` `=("Mass of "HCl)/("Molar mass of "HClxx("Weight of solution")/("density of solution"xx1000))` `=(38)/(36.5xx(100)/(1.19xx1000))=12.4` (b) MIlli-moles of solute does not change on dilution and this, `{:(MxxV_(mL),=,MxxV_(mL),,),("(for conc.)",=,"(for dil)",,):}` `12.4xxV= 0.1xx100 ` `V= 8.06mL` |
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| 42. |
For the dissolution of `1.08 g` of metal, `0.49 g` of `H_(2)SO_(4)` was required. If specific heat of metal is `0.06 cal//g`, what is its atomic mass? |
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Answer» App. At. Wt. `= (6.4)/(0.06) = 10.6.67` Let Eq. wt. of metal be `E`. `(1.08)/(E) = (0.49)/(49)` `:. E_("metal") = 108` Also valence `= ("at.wt.")/("Eq. wt.") = (106.67)/(108) = 1` (integer) `:.` Exact at wt. `= "Eq. wt" xx " valence" = 108 xx 1 = 108` |
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| 43. |
Determine the equivalent weight of each given below, if formula weight of these compounds are `X,Y` and `Z` respectively: (i) `Na_(2)SO_(4)`, (ii) `Na_(3)PO_(4). 12H_(2)O` (iii) `Ca_(3)(PO_(4))_(2)` |
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Answer» Equilibrium weight of compound `= ("Molecular weight")/("Total charge on cartion or anion")` `E_(Na_(2)SO_(4)) = (x)/(2)` (charge on `2Na` is `+2` or on `SO_(4) is -2)` `E_(Na_(3)PO_(4).12H_(2)O) = (y)/(3)` , (charge on `PO_(4)` is `-3`) `E_(Ca_(3)(PO_(4))_(2) = (z)/(6)` , (charge on `2PO_(4)` is `-6`) |
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| 44. |
Water contaminated with `H_(2)S` can be freed from `H_(2)S` by passing `Cl_(2)` through it. If the `H_(2)S` content in contaminated water is `22` ppm by mass how much `Cl_(2)` is needed to remove all the `H_(2)S` from `2xx10^(2)` gallons of water. `(1 gallon =3.785 litre)` |
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Answer» `because 10^(6)g` or `mL H_(2)O` contain `22g H_(2)S` `:. 2xx 10^(2) xx 3.785 xx 10^(3) xx 10^(3) mL` `= (22 xx 2 xx 10^(2) xx 3.785 xx 10^(3))/(10^(6))` `= 22 xx 2 xx 3.785 xx 10^(-1)` `= 16.654gH_(2)S` `Cl_(2)(g)+H_(2)S(aq).rarr2HCl_(aq)+S(s)` Eq. of `Cl_(2) =` Eq. of `H_(2)S` `(w)/(35.5) = (16.654)/(34//2)` `w = 34.78 g Cl` |
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| 45. |
Calculate the degree of hardeners of river water whose `100 mL` solution required `1.68 mL` of `0.1 N H_(2)SO_(4)`. |
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Answer» Meq. Of `H_(2)SO_(4)` used `= 1.68 xx 0.1 = 0.168` Assuming `CaCO_(3)` in water, Meq. Of `CaCO_(3) = 0.168` `:.(w)/(50) xx 1000 = 0.168` `:. w_(CaCO_(3)) = 8.4 xx 10^(-3) g` `because 100 mL` water `-= 8.4 xx 10^(-3)g CaCO_(3)` `:. 10^(6) mL` water `-= (8.4 xx 10^(-3) xx 10^(6))/(100)` `= 8.4 xx 10 = 84`ppm |
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| 46. |
Calculate the volume of 0.5 M `H_2SO_4` required to dissolve 0.5 g of copper (II) carbonate `(CuCO_3)`. |
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Answer» Meq.of `H_(2)SO_(4)` Meq. of `CuCO_(3)` `("Eq. wt. of" CuCO_(3) = (M)/(2))` ltbr. `0.5 xx 2 V = (0.5 xx 2 xx 1000)/(1235)` `:. V = 8.097 mL` |
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| 47. |
Calculate the volume of 0.5 M `H_2SO_4` required to dissolve 0.5 g of copper (II) carbonate `(CuCO_3)`.A. `8.10 mL`B. `16.20 mL`C. `4.05 mL`D. `12.05 mL` |
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Answer» Correct Answer - A `"Meq.of"H_(2)SO_(4)="Meq.of"CuCO_(3)` `0.5xx2xxV=(0.5xx1000)/(123.5//2)` `("mol.wt".CuCO_(3)=123.5)` `:. V=8.1 mL` |
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| 48. |
Find the weight of `H_(2) SO_(4)` in `1200 mL` of a solution of `0.2 N` strength.A. `11.76 g`B. `12.76 g`C. `13.76 g`D. `23,52 g` |
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Answer» Correct Answer - A `Meq. of H_(2)SO_(4)=1200xx0.2` `(w)/(49)xx1000=1200xx0.2` `:. w_(H_(2)SO_(4))=11.76 g` |
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| 49. |
Volume strength of `H_(2)O_(2)` labelled is `10 vol`. What is normality of `H_(2)O_(2)`?A. `1.79`B. `12.79`C. `0.79`D. `5.6` |
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Answer» Correct Answer - A `"Normality"=("Volume strength of"H_(2)O_(2))/(5.6)` `:. N=(10)/(5.6)=1.79` |
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| 50. |
Which does not change on dilution?A. Molarity oof solutionB. Molality of solutionC. Milli-moles and milli-equivalent of solutionD. Mole fraction of solute |
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Answer» Correct Answer - C Moles, equivalent of solute do not change n dilution. |
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