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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.
| 301. |
A 36 mL mixture of an alkene and propane required 171 " mL of " `O_2` for complete combustion and yielded 109 " mL of " `CO_2` (all volume measured at same temperature and presasure). Calculate the molecular formula of olefin and composition of the mixture by volume. |
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Answer» `x " mL of " C_(n)H_(2n)` `(36-x)" mL of " C_3H_8` `C_(n)H_(2n)+(3n)/(2)O_2tonCO_2+nH_2O` `C_(3)H_(8)+5O_2to3CO_2+4H_2O` Volume of `CO_2=nx+3(36-x)=108` `impliesn=3 or x=0` (impossible) Volume of `O_2` used`=171` mL `thereforex((3n)/(2))+5(36-x)=171` Substituting `n=3,ximplies18mL` The hydrocarbon is `C_3H_6` and the mixture is `50%` composition by volume because `x=18mL` |
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| 302. |
A 20 mL mixture of ethane, ethylene, and `CO_2` is heated with `O_2`. After the explosion, there was a contraction of 28 mL and after treatment with KOH, there was a further contraction of 30 mL. What is the composition of the mixture? |
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Answer» Let the volume of ethane, ethylene, and `CO_(2)`, respectively, are a,b and `(20-a-b)mL` Now contraction after cooling`=28mL` `thereforeV_(R)-V_(P)=25` `V_(R)=` Volume of `C_(2)H_(6)+` volume of `C_(2)H_(4)+` volume of `CO_(2)+` volume of `O_(2)` used for combustion `V_(P)=` volume of `CO_(2)` produced (Volume of `H_(2)O` is not taken since it condenses) Considering the combustion gases. (a). `C_(2)H_(6)+(7)/(2)O_(2)to2CO_(2)+3H_(2)O` (b). `C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O` (c). `CO_(2)to` no reaction `V_(R)=(a+(7)/(2)a)+(b+3b)+(20-a-b)` `V_(P)=(2a+2b)+(20-a-b)` `V_(P)-V_(R)=(5)/(2)a+2b=28` `therefore5a+4b=56` ..(i) There is a further contraction of 32 mL on treatment with KOH. Volume of `CO_(2)` produced `+` Volume of `CO_(2)` original `=32` `(2a+2b)+(20-a-b)=32` `therefore a+b=12` ...(ii) On solving (i) and (ii) we get `a=8mL` (volume of `C_(2)H_(6))` `b=4mL` (Volume of `C_(2)H_(4))` Volume of `CO_(2)=8mL` |
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| 303. |
A mixture of CO and `CO_2` having a volume of 30 mL is mixed with x " mL of " `O_2` and electrically sparked. The volume after explosion is `(20+x)` mL under the same condition. What would be the residual volume if 45 " mL of " the original mixture is treated with aueous NaOH? (a). 10 mL (b). 20 mL (c). 30 mL (d). mL |
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Answer» `CO(g)+(1)/(2)O_2(g)toCO_2(g)` `"Final"-(x-(y)/(2))mL ymL` `CO_2(g)+O_2(g)to` no reaction `underline("Total volume of gases left"=(30-y)+(x-(y)/(2))+y=(20+x))` `thereforey=20mL` In `45 mL, V_(CO)=(20)/(30)xx45=30mL` and `V_(CO_2)=45-30=15mL` Only `CO_2` reacts with `NaOH`, so `15 " mL of " CO_2` reacts. `therefore` reasidual volume `V_(CO)=30mL` |
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| 304. |
25 " mL of " a mixture of Co, `CO_(2)` and `H_(2)` were exploded mL with 10 " mL of " oxygen. The products has a volume of 18.5 " mL of " which 17 mL absorbed by alkali. What was the composition of the original mixture? All volume measurements were made at the same temperature and pressure? |
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Answer» Let `V_(1)` mL, `V_(2)mL` and `V_(3)mL` be the volumes of `CO,CO_(2)` and `H_(2)` respectively. Then `V_(1)+V_(2)+V_(3)=25` Contraction after explosion`=(25+10-18.5)mL` `=16.5mL` `(V_(1))/(2)+(3V_(3))/(2)=16.5` Absorption `-=17mL=V_(1)+V_(2)`. Solving these, we have `V_(3)=8mL`, `V_(1)=9mL and V_(2)=8mL` `thereforeCO:CO_(2):H_(2)=9:8:8` |
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| 305. |
3.0 g of metal chloride gave 2.0 g of metal. Calculate the equivalent weight of the metal. |
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Answer» Let the equivalent weight of metal (M) be E. `Ew(Cl)=35.5g`. Weight of `Cl=3.0-2.0=1.0g` " Eq of "mwetal `=" Eq of "`Cl ltBrgt `("Weight of M")/(Ew(M))=(Weight of Cl)/(Ew(Cl))` `(2.0)/(E)=(1.0)/(35.5)` `E=71.0g` General formula: If x g of metal chloride gave y g of metal, the Ew of metal is Ew of `M` `=("weight of M")/("weight of M chloride"-"weight of M")xxEw(Cl)` `=(x xx35.5)/(x-y)=(2.0xx35.5)/((3.0-2.0))=71.0g` |
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| 306. |
What volume of 0.25 `MH_2SO_4` is required to neutralise 1.90 g of a mixture containing equimolar amounts of `NaHCO_3` and `NaCO_3`? |
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Answer» x mol each of `NaHCO_3` and `Na_2CO_3` `implies84x+106x=1.9g` (given) `x=0.01` `implies` milli" mol of "each `=10` Now m" Eq of "`H_2SO_4=m" Eq of "(NaHCO_3+Na_2CO_3)` `2xx0.25xxV=1xx10+2xx10` `impliesV=60mL` |
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| 307. |
`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. `(2)/(3) " mol of "MnO_(4)^(ɵ) and (1)/(3) " mol of "MnO_(2)`B. `(1)/(3) " mol of "MnO_(4)^(ɵ) and (2)/(3) " mol of "MnO_(2)`C. `(1)/(3) " mol of "Mn_(2)O_(7) and (2)/(3) " mol of "MnO_(2)`D. `(2)/(3) " mol of "Mn_(2)O_(7) and (1)/(3) " mol of "MnO_(2)` |
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Answer» Correct Answer - A Remember it as fact: in nautral medium `K_(2)MnO_(4)` (potassium manganate) disproportionate as follows: `K_(2)MnO_(4)toKMnO_(4)+MnO_(2)` On balancing we get: `3MnO_(4)^(2-)+2H_(2)Oto2MnO_(4)^(ɵ)-=(1)/(3)molMnO_(2)` |
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| 308. |
The mass of one litre sample of ozonised oxygen at `NTP` was found to be `1.5 g`. When `100 mL` of this mixture at `NTP` were treated with terpentine oil, the volume was reduced to `90 mL`. Hence calculate the molecular mass of ozone. (Terpentine oil absorbs ozone) |
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Answer» Volume of ozone=Volume absorbed in turpentine oil =10mL Volume of oxygen=90mL Mass of 100mL mixture `=(10)/(22400)xxM+(90)/(22400)xx32` On solving, we get M=48 Molecular mass of ozone=48 |
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| 309. |
The weight of 1 L sample of ozonised oxygen at STP was found to be 1.5 g When 100 " mL of " this mixture at STP was treated with turpentine oil, the volume was reduced to 90 mL. Calculate the molecular weight of ozone. |
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Answer» Volume absorbed by turpentive oil`=10mL` `=` volume of `O_(3)` Volume of `O_(2)=100-10=90mL` Mw of ozonised oxygen`=(WeightxxRT)/(PV)` `=(1.5xx0.0821xxRT)/(1xx1)=33.62` Volume of mol ratio of `O_(2)` and `O_(3)` `=900:100` Mw of ozonised oxygen`=(900xx32+100xxa)/(1000)` `33.62=(900xx32+100xxa)/(1000)` or `a=48.2` Mw of `O_(3)=48.2` |
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| 310. |
1.53 g of a compound containing only sulphur, oxygen and chlorine after easy hydrolysis with water yielded acid products which consumed 91 " mL of " `(N)/(2)` sodium hydroxide for complete neutralisation in a parallel experiment, 0.4 g of the compound after hydrolysis with water, was treated with excess of `BaCl_(2)` solution and 0.7 g of `BaSO_(4)` was precipitated. What is the formula of the compound? |
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Answer» After hydrolysis, acid products are obtained. This suggests that the substanece is an acid chloride. Reaction with `BaCl_(2)` to yield `BaSO_(4)` implies that `H_(2)SO_(4)` is one of the product. The substance is (by surmise) `SO_(2)Cl_(2)` This can be verified by the following way: `SO_(2)Cl_(2)(135g)=BaSO_(4)(233.4g)` `therefore0.4g-=[(233.4)/(135)xx0.4]g=0.692g` This agrees very nearly with the given data `=0.7g` further, 135 g of `SO_(2)Cl_(2)-=H_(2)SO_(4)+2HCl` (by hydrolysis 4 equivalents). `1.53g-=(4)/(135)xx1.53` equivalent `=0.0453` equivalent `91 " mL of " 0.5 N NaOH-=(91)/(1000)xx0.5=0.0455` equivalent This also agrees with the given data. Therefore, the formula is `SO_(2)Cl_(2)`. |
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| 311. |
When `1xx10^(-3)` " mol of "the chloride of an elements Y was completely hydrolysed, it was found that the resulting wsolution requried 20 " mL of " 0.1 M aqueous silver nitrate for complete precipitation of the chloride ion. Elements Y could beA. AluminiumB. PhosphorusC. SiliconD. Sulphur |
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Answer» Correct Answer - D `Ycl_(n)+nH_(2)OtonHCl+Y(OH)_(n)` `(n=` valency of `Y`) `nHCl+nAgNO_(3)tonAgCl` `(1)/(n)=(1xx10^(-3))/(20xx0.1xx10^(-3))` `((Molaityxxvolume)/(1000)=mol)` `n=2` (other have `ngt2`) |
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| 312. |
56.0 g KOH, 138.0 g `K_(2)CO_(3)` and 100.0 g `KHCO_(3)` is dissolved in water and the solution is made 1 L. 10 " mL of " this stock solution is titrated with 2.0 M HCl. Which of the following statements is/are correct?A. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 60 mLB. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 40 mL.C. When methyl orange is used as an indicator from the very beginning, the titre value of HCl will be 80 mL.D. When methyl orange is used as an indicator after the first end point the titre value of HCl will be 30 mL. |
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Answer» Correct Answer - B::C::D `m" Eq of "`(KOH)/(L)=(56)/((56)/(1))xx10^(3)=100=(20)/(20)mL` solution m" Eq of "`(K_(2)CO_(3))/(L)=(138)/((138)/(2))xx10^(3)=200=(40)/(20)mL` solution `m" Eq of "(KHCO_(3))/(L)=(100)/((100)/(1))xx10^(3)=1000=(20)/(20)mL` (b). With phenolphthalein: `m" Eq of "KOH+(1)/(2)m" Eq of "K_(2)CO_(3)=m" Eq of "HCl` `20+(40)/(2)=vxx1N(n=1)` `V_(HCl)=40mL` (c). With methyl orange: `m" Eq of "KOH+m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` ltBrgt `20+40+20=Vxx1N` `V_(HCl)=80mL` (d). With methyl orange after the first end point: `(1)/(2)m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` `20+20=Vxx1N` `V_(HCl)=40mL` |
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| 313. |
Magnetite, `Fe_(3)O_(4)`, can be converted into metallic iron by heating with carbon monoxide as represented by this equation: `Fe_(3)O_(4)(s)+CO(g) rarr Fe(s) +CO_(2)(g)` The kilograms of `Fe_(3)O_(4)` which must be processed in this way to obtain `5.00 kg` of iron, if the process is `85%` efficient is closest to? `[M: = Fe = 56]`A. 6.92kgB. 8.12kgC. 20.8kgD. 24.4kg |
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Answer» Correct Answer - B `1"mole "Fe_(4)O_(4)(232g)=3"mole Fe" (168g)` Amounts of `Fe_(3)O_(4)` required for 5kg iron=`(232)/(168)xx5kg` =6.904kg Since, efficiency of the reaction si 85% hene, the actual required amount of `Fe_(3)O_(4)` will be `=(100xx6.904)/(85)kg i.e.=8.12kg` |
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| 314. |
A 1.0g sample of a pure organic compound cotaining chlorine is fused with `Na_(2)O_(2)` to convert chlorine to NaCl. The sample is then dissolved in water, and the chloride precipitated with `AgNO_(3)`, giving 1.96 g of AgCl. If the molecular mass of organic compound is 147, how many chlorine does each molecule contain ?A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B |
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| 315. |
A 0.6gm sample consisting of only `CaC_(2)O_(4)` and `MgC_(2)O_(4)` is heated at `500^(@)C`gets converted into `CaCO_(3)` and `MgCO_(3)`. The sample then weighed 0.465gm. If the sample had been heated to `900^(@)C` where the products are CaO and MgO, then what would the mixture of oxides weigh?A. 0.12 gB. 0.21 gC. 0.252 gD. 0.3 g |
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Answer» Correct Answer - C |
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| 316. |
638.0 g of `CuSO_(4)` solution is titrated with excess of 0.2 M KI solution. The liberated `I_(2)` required 400 " mL of " 1.0 M `Na_(2)S_(2)O_(3)` for complete reaction. The percentage purity of `CuSO_(4)` in the sample isA. `5%`B. `10%`C. `15%`D. `20%` |
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Answer» Correct Answer - B `CuSO_(4)+KItoI_(2)toS_(2)O_(3)^(2-)` `2Cu^(2+)+2e^(-)toCu_(2)^(o+)` `underline(2I^(ɵ)toI_(2)+2e^(-))` `underline(2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2))` `I_(2)+2S_(2)O_(3)^(2-)to2I^()+S_(4)O_(6)^(2-)` `2 mol CuSO_(4)-=4 mol KI-=1 mol I_(2)` `-=2 mol S_(2)O_(3)^(2-)` `mmol CuSO_(4)-=mmol S_(2)O_(3)^(2-)` `-=400xx1mmol=0.4mol` Weight of `CuSO_(4)-=400xx10^(-3)xx159.5g` `% CuSO_(4)=(400xx10^(-3)xx159.5xx100)/(638)=10%` |
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| 317. |
1.0 g a of moist sample of a mixture of KCl and `KClO_3` was dissolved in water and made up to `250mL`. 25 " mL of " this solution was treated with `SO_2`. The chlorate was reduced to chloride and excess of `SO_2` was removed by boiling. The total chloride was precipitated as `AgCl` . The weight of the precipitate was `0.1435 g`. In another experiment, 25 " mL of " the original solution was heated with 30 " mL of " 0.2 N solution of ferrous sulphate, and the unreacted ferrous sulphate required 37.5 " mL of " 0.08 N solution of an oxidising agent for complete oxidation. Calculate the molar ratio of the chlorate to the chloride in the given mixture `Fe^(2+)` reacts with `ClO_3^(ɵ)` according to the equation. `ClO_3^(ɵ)+6Fe^(2+)+6H^(o+)toCl^(ɵ)+6Fe^(3+)+3H_2O` |
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Answer» Total `Cl^(ɵ)` (from `KCl+KClO_3)(ClO_3^(ɵ)toCl^(ɵ))` 25 " mL of " the mixture`=0.1435 g of AgCl` `=(0.1435)/(143.5)` " mol of "AgCl `=10^(-3) " mol of "Cl^(ɵ)` ions Second experiment: Total volume of `FeSO_4=30mL` Excess volume of `FeSO_4=(37.5xx0.08)/(2)=15mL` Volume of `FeSO_4` used `=30-15=15mL` `15 mL` of `0.2N FeSO_4` used up `=15xx0.2xx10^(-3) " Eq of "FeSO_4` `=3xx10^(-3) " mol of "FeSO_4` Since 1 mol `ClO_3^(ɵ)` uses 6 " mol of "`FeSO_4`, `3xx10^(-3)` " mol of "`FeSO_4=(1)/(6)xx3xx10^(-3)` `=0.5xx10^(-3) " mol of "ClO_3^(ɵ)` `Cl^(ɵ)+ClO_3^(ɵ)=10^(-3)` `thereforeCl^(ɵ)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)` `thereforeCl^(ɵ)=0.5xx10^(-3),ClO_3^(ɵ)=0.5xx10^(-3)` |
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| 318. |
A metal M forms the sulphate `M_(2)(SO_(4))_(3)`. A 0.596 gram sample of the sulphate reacts with excess `BaCl_(2)` to give 1.220 g `BaSO_(4)`. What is the atomic mass of M ?A. 26.9B. 69.7C. 55.8D. 23 |
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Answer» Correct Answer - A |
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| 319. |
5g of `K_(2)SO_(4)` are dissolved in 250mL of solution. How many mL of this solution should be used so that 1.2 of `BaSO_(4)` may be precipated from `BaCl_(2)` solution? |
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Answer» The desired equation is `BaCl_(2)+underset(=174g)underset(2xx39+32+64)(K_(2)SO_(4)) to underset(=223g)underset(137+32+64)(BaSO_(4))+2KCl` `233g of BaSO_(4)` obtained from 174g of `K_(2)SO_(4)` 1.2 of `BaSO_(4)` will be obtained from `(174)/(233)xx1.2` `=0.8961"g of "K_(2)SO_(4)` So, 0.891g of `K_(2)SO_(4)` will be present in `(250)/(5)xx0.8961` `=44.8"mL of solution"` |
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| 320. |
25mL of `2N HCl, 50 mL "of" 4N HNO_(3) and xmL H_(2)SO_(4)` are mixed together and the total volume is made up to 1L after dilution. 50mL if this acid ixture completely reacteed with 25mL of a `1 N Na_(2)CO_(3)` solution. The value of x is:A. 250mLB. 62.5mLC. 100mLD. none of these |
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Answer» Correct Answer - B |
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| 321. |
What volume of `3M Na_(2)SO_(4)` must be added to 25mL of 1M `BaCl_(2)` to produce `5g BaSO_(4)`? |
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Answer» Correct Answer - a `Na_(2)SO_(4)+BaCl_(2) to BaSO_(4)+2NaCl` `"Number of moles of "BaSO_(4)=("Mass")/("Molar mass")=(5)/(233)=0.0215` `(MV)/(1000)=0.0215 ` `(3xxV)/(1000)=0.0215` V=7.2mL |
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| 322. |
25mL of 0.15M Pb `(NO_(3))_(2)` reacts completely with 20 mL of `Al_(2)(SO_(3))_(2)`. The molar concentration of `Al_(2)(SO_(4))_(3)` will be: `3Pb(NO_(3))_(2)(aq.)+Al_(2)(SO_(4))_(3)(aq.)to3PbSO_(3)(s)+2Al(NO_(3))_(3)(aq.)` |
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Answer» Correct Answer - a `(M_(1)V_(1))/(n_(1))Pb(NO_(3))_(2)=(M_(2)V_(2))/(n_(2))Al_(2)(SO_(4))_(3)` `(0.15xx25)/(3)=(M_(2)xx20)/(1)` `M_(2)`=0.0625 |
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| 323. |
1.25g of a solid dibasic acid is completely neutralised by 25mL of 0.25 molar `Ba(OH_(2))` solution. Molecular mass of the acid is:A. 100B. 150C. 120D. 200 |
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Answer» Correct Answer - D |
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| 324. |
What volume of `NH_(3)(g) at 27^(@)C` and 1atm pressure will be obtained by therma, decomposition of 26.25g `NH_(3)Cl`? |
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Answer» Amminomium chloride undergoes decomposition as, `underset(53.5g)underset(1"mol")(NH_(4)Cl(s)) to underset(1"mol")underset(1"mol")(NH_(3)(g))+HCl(g)` `therefore 53.5g NH_(4)Cl "give 1 mol"NH_(3)` `=0.5"mole"` `Pv=Nrt` `1xxV=0.5xx0.0821xx300` V=12.315litre |
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| 325. |
If 1 mole of `H_(3)PO_(4)` reacts with 1 mole of `X(OH)_(2)` as shown below : `H_(3)PO_(4)+X(OH)_(2) to XHPO_(4)+2H_(2)O " then"`A. the equivalent mass of base is `(mol.mass)/(2)`B. the eq. mass of `H_(3)PO_(4)` is `(98)/(3)`C. the resulting solution requires 1 mole NaOH for complete neutralizationD. minimum 1 mole of X`(OH)_(2)` is required for complete neutralization of `XHPO_(4)` |
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Answer» Correct Answer - A::C |
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| 326. |
On dissolving 2.0 g of metal in sulphuric acid ,4.51 g of the metal sulphate was formed . The specific heat of the metal is 0.057 cal `g^(-1).^(@)C^(-1)`. What is the valency of metal ? |
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Answer» Correct Answer - 3 |
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| 327. |
2.1 g of a mixture of `NaHCO_3` and `KCIO_3` required 100 " mL of " 0.1 HCl for complete reaction. Calculate the amount of residue that would be obtained on heating 2.2 g of the same mixture strongly. |
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Answer» `Ew(NaHCO_3)=23+1+12+3xx16=(84)/(1)=84(n=1)` HCl reacts only with `NaHCO_3` (acid and base reaction) `m" Eq of "HCl=100xx0.1` `=10m" Eq of "NaHCO_3` `=10xx10^(-3)xx84g of NaHCO_3=0.84g` Weight of `KClO_3=2.1-0.84=1.26g` Weight of `NaHCO_3` in `2.2 g of mixture=(0.84xx2.2)/(2.1)` `=0.88g` Weight of `KClO_3` in 2.2 g of mixture `=(2.2-0.88)g` `=1.32`g Heating of mixture: `NaHCO_3overset(triangle)toNa_2CO_3+H_2O+CO_2` Weight of residue obtained on heating `NaHCO_3` `=(106xx0.88)/(2xx84)=0.555g` `2KClO_3to2KCl+3O_2` Weight of residue obtained on heating `KClO_3` `=(2xx74.5xx1.32)/(2xx122.5)=0.802g` Total weight of residue `=0.555+0.802=1.357g` |
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| 328. |
Oxidation number of `P` in `Ba(H_(2)PO_(2))_(2)` isA. `-1`B. `+1`C. `+2`D. `+3` |
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Answer» Correct Answer - B |
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| 329. |
Calculate the total moles of atoms of each element present in 122.5 g of `KC LO_(3)` |
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Answer» Correct Answer - 5 |
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| 330. |
The composition of a sample of wustite is `Fe_(0.93)O_(1.00)`. What percentage of iron is present in the form of `Fe(III)`? |
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Answer» First method: In pure iron oxide, Fe and O are in the ratio of `1:1`. The compound `Fe_(0.93)O_(1.0)` as a whole should remain neutral. Suppose x atoms of `Fe^(3+)` ions are present in the compound of `Xfe^(3+)` atoms have replaced n atoms of `Fe^(2+)`. Number of atoms of `Fe^(2+)=0.93-x` For neutrality of compound. Total positive charge `=` Negative charge on compound, `implies` charge on iron atom`=`charge on O atom or charge on `(Fe^(2+)+Fe^(2+))=2` `2(0.93-x)+3x=2` `1.86+x=2` `x=0.14` `%` of iron present as `Fe^(3+)=(0.14)/(0.93)xx100=15.05%` Second method: Oxidation method of Fe in `Fe_(0.093)O=0.93xx x-2xx1=0`, `x=2.15`. Oxidation number of Fe is an intermediate value of `Fe^(2+)` and `Fe^(3+)` Let `%` of `Fe^(3+)=a` `(3xxa+2(100-a))/(100)=2.15` `thereforea=15.05%` `%` of `Fe^(3+)=15.05` Third method: Let x atom of `Fe^(3+)` ions be present in the compound. `F_(0.93)` (number of `Fe^(3+)` ions of `Fe^(2+)` ions.) Number of `Fe^(2+)` ions `=` Total `-` number of `Fe^(3+)=(0.93-x)` Total positive charge on `Fe^(2+)+Fe^(3+)=` Total negative charge on oxygen. `2(0.93-x)+3x=2,becausex=0.14` `therefore% of Fe^(3+)=(0.14)/(0.93)xx100=15.05` Third Method: Let x atoms of `Fe^(3+)` ions be present in the compound `F_(0.93)` (number of `Fe^(3+)` ions of `Fe^(2+)` ions) Number of `Fe^(2+)` ions=total-number of `Fe^(3+)=(0.93-x)` total positive charge on `Fe^(2+)+Fe^(3+)=` total negative charge on oxygen `2(0.93-x)+3x=2,therefore0.14` `therefore%` of `Fe^(3+)=0.14//0.93xx100=15.05` |
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| 331. |
W is the mass of iron (in g) which will be converted into `Fe_(3)O_(4)` by the action of 18 g of steam on it . What is the value of W/7 ? `Fe+H_(2)O to Fe_(3)O_(4)+H_(2)` |
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Answer» Correct Answer - 6 |
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| 332. |
`4.08 g` of a mixture of `BaO` and an unknown carbonate `MCO_(3)` was heated strongly. The residue weighed `3.64 g`. This was dissolved in `100 mL` of `1 N HCl`. The excess of acid required of `16 mL` of `2.5 N NaOH` for complete neutralisation. Identify the metal `M`. |
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Answer» Weight of mixture of BaO and `MCO_3=4.08g` On heating of `MCO_3` gives MO and `CO_2` Weight of `BaO+MO=3.64g` Weight of `CO_2` evolved `=4.08-3.64=0.44g` Mols of `CO_2=(0.44)/(44)=10^(-2)mol=10^(-2)xx10^(3)` `=10m mol =20 m" Eq of "CO_2` `=20mEq MCO_3` Total acid `=100xx1=100mEq` Excess of acid`=`Total acid`-`m" Eq of "`NaOH` `=100-16xx2.5=60mEq` of excess acid `(Ew(BaO)=(Mw)/(2))` `m" Eq of "BaO=60-20=40` mEq `=((138+16)xx40)/(100xx2)=3.08g` weight of `MO=3.64-3.08=0.56g` 20 m" Eq of "`MO=0.56` Equivalent weight of `MO=0.56xx1000=28` Molecular weight of `MO=28xx2=56` Hence, atomic weight of `M=56-16=40`. The metal is |
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| 333. |
The volume strength of `1.5 N H_2O_2` solution isA. 4.8B. 8.4C. 3D. 8 |
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Answer» Correct Answer - B `underset(68g)2H_(2)O_(2)tounderset(22.4 L)(2H_(2)O+O_(2)` Mass opf `H_(2)O_(2)` in 1.5 N solution`=Ew of H_(2)O_(2)xx1.5N` `=17xx1.5=25.5g` So, volume strength of 1.5 `H_(2)O_(2)` solution `=(22.4Lxx25.5g)/(68.0)=8.4L` |
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| 334. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(ɵ)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - B `MnSO_(4)` `x+6-8=0, x=2` In (a) `Mn_(2)O_(3), 2x-6=0, x=3` in (b) `MnO_(2), x-4=0, x=4` In (c) `MnO_(4)^(ɵ), x-8=-1,x=7` In (d) `MnO_(4)^(2-), x-8=-2,x=6` In (b), the change in oxidation number is `4-2=2` Therefore, the answer is (b). |
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| 335. |
2 g of `FeC_(2)O_(4)` are made to react in acid solution with `0.25 M KMnO_(4)` solution. What volume of `KMnO_(4)` would be required? The be reqruied? The resulting solution is treated with excess of `NH_(4)Cl` solution and `NH_(4)OH` solution. The precipitated `Fe(OH)_(3)` is filtered off, washed and ignited. What is the mass of the product obtained? (Atomic weight of `Fe=56`) |
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Answer» `FeC_(2)O_(4)(144g)-=1+2=3` equivalent `2 g FeC_(2)O_(4)-=(2)/(144)xx3=(1)/(24)` equivalent `0.25 M KMnO_(4)-=1.25N` `(V)/(1000)xx1.25=(1)/(24)impliesV=(1000)/(24xx1.25)=33.33 mL` i.e., volume of `KMnO_(4)` required `=33.33mL` Weight of Fe in the sample is `(56)/(144)xx2=(112)/(144)g`. Since `underset(112g)(2Fe)+3(O)tounderset(160g)(Fe_(2)O_(3))` `(112)/(144)g-=(160)/(112)xx(112)/(144)=1.11g` i.e., the mass of product obtained`=1.11g` |
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| 336. |
Two drops of phenolphthalein was added to 40 " mL of " HCl solution. When 30 " mL of " 0.1 M NaOH was added, part of the the solution turned pink, but colour disappeared on mixing the solutiion. Addition of NaOH was continued drop-wise untill a one-drop addition produced a lasting pink colour, and the colume of NaOH added was 32.56 mL. Calculate (a). The concentration of HCl solution. (b). The concentration of HCl solution when 30 mL base was added. (c). The pH of solution when 30 mL base was added. (d). The pH of solution when 32.56 mL base was added |
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Answer» (a). At the end point: `m" Eq of "HCl=m" Eq of "NaOH` `40xxN=01xx1xx32.56` `N_(HCl)=0.0814` (b). `m" Eq of "HCl=0.0814xx40=3.256` `m" Eq of "NaOH added =30xx0.1=3.0` `m" Eq of "HCl` left`=3.256-3.0=0.256` `N_(HCl)=(m" Eq of "HCl)/("Total volume")=(0.256)/(30+40)` `=3.66xx10^(-3)` `thereforeN_(HCl)=3.66xx10^(-3)` (c). `pH=-log[H^(o+)]=-log(3.66xx10^(-3))=2.4365` (d). At the end point `pH=7`. |
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| 337. |
Calculate the volume strength of `H_2O_2` solution if 50 " mL of " this diluted solution required 40 " mL of " `(M)/(60)` `K_2Cr_2O_7` solution in presence of `H_2SO_4` for complete reaction. |
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Answer» `H_2O_2-=Cr_2O_7^(2)` `(n=2)(n=6)` `mEq-=mEq` `N_1V_1-=N_2V_2` `N_1xx20-=(1)/(60)xx6xx40` `N(H_2O_2)=(4)/(20)=0.2N` The normality of the 50 " mL of " `H_2O_2` should be twice the normality of 20 " mL of " the diluted solution of `H_2O_2`. Since the volume of `H_2O_2` solution has doubled. `(50 " mL of " H_2O_2+50" mL of " H_2O=` 100 " mL of " solution of `H_2O_2`) `N(H_2O_2) of 50 " mL of " H_2O_2=0.2xx2=0.4N` Volume strength of `H_2O_2=5.6xx0.4=2.24` volume of `O_2` |
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| 338. |
Calculate the weight of `K_2Cr_2O_7` required to produce from excess oxalic acid `(H_2C_2O_4),8.2L` `CO_2` at `127^2` and 1.0 atm pressure? |
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Answer» `underset((1 mol))(Cr_2O_7^(2-))+8H^(o+)+3H_2C_2O_4underset((6 mol))(6CO_2)+2Cr^(3+)+7H_2O` `n=(PV)/(RT)=(1 atmxx 8.2L)/(0.082L" atm mol"^(-1)xx400K)` `=0.25molCO_2` `Mw(K_2Cr_2O_7)=2xx39+2xx52+16xx7=294g` 6 mol `CO_2` is obtained from 1 " mol of "`K_2Cr_2O_7` 0.25 mol `CO_2` is obtained `=(1xx0.25)/(6)=0.04` mol `=0.04xx294g=11.76g` |
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| 339. |
A mixture of formic acid and oxalic acid is heated with conc `H_2SO_4`. The gas produced is collected and treated with KOH solution, where the volume decreases by `(1)/(6)` The molar ratio of the two acids (formic acid/oxalic acid) in the original is (a). `4:1` (b). `1:4` (c). `2:1` (d). `1:2` |
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Answer» Let x " mol of "HCOOH and y " mol of "`(COOH)_2` be taken `underset(x)(HCOOH)overset(H_2SO_4)toH_2O+underset(x)(CO)` `underset(y)((COOH)_2)overset(H_2SO_4)toH_2Ounderset(y)(CO)+underset(y)(CO_2)` Total " mol of "`CO+CO_2=x+2y` Total " mol of "`CO_2=y` According to the question `(x+2y)xx(1)/(6)=y` `therefore(x)/(y)=4:1` `therefore` Alternatively Mol fraction of `CO_2=(y)/(x+2y)=(1)/(6)` `therefore(x)/(y)=4:1` |
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| 340. |
A mixture of `H_2C_2O_4` and `HCOOH` is heated with conc `H_2SO_4`. The gas produced is collected and on treatment with KOH solution, the volume of the gas decreases by `(1)/(6)` calculate the molar ratio of the two acids in the original mixture. |
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Answer» `H_2C_2O_4underset(triangle)overset(H_2SO_4)toH_2O+underset(y)(CO)+underset(y)(CO_2)` `underset(x)(HCOOH)underset(triangle)overset(H_2SO_4)toH_2O+underset(x)(CO)` Let x and y moles of HCOOH and `H_2C_2O_4` be present in the original mixture. Total moles of CO formed `=x+y` Moles of `CO_2` formed `=y` Total moles of gases `=x+y+y=x+2y` Since KOH solution absorbs `CO_2` and volume reduces by `(1)/(6)`. Moles of `CO_2(1)/(6)(x+2y)` `y=(1)/(6)(x+2y)` `(x)/(y)=4` `x:y::4:1` |
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| 341. |
Solutions containing 23 g HCOOH is/are :A. `46 g " of " 70%((w)/(V))HCOOH(d_("solution" )=1.40g//mL)`B. `50 g " of 10 M " HCOOH(d_("solution" )=1g//mL)`C. `50 g " of " 25 % ((w)/(w)) HCOOH`D. 46 g " of 5 M " HCOOH`(d_(solution)=1 g//mL)` |
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Answer» Correct Answer - A::B |
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