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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.
| 251. |
Calculate the volume of hydrogen liberated at `27^(@)C and 760mm` pressure by treating 1.2g of magnesium with excess of hydrochloric acid. |
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Answer» The balanced equation is `underset(24g)underset(1"mol")(Mg)+2HCl to MgCl_(2)+underset(22.4"litre of NTP")underset(1"mol")(H_(2))` 24g of Mg liberates hydrogen=22.4litre 1.2g Mg will libertes hydrogen=`(22.4)/(24)xx1.2=1.2"litre` Volume of hydrogen under give condition can be calculated by applying `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `{:(,P_(1)=760mm,P_(2)=760mm),(,T_(1)=273K,T_(2)=(27+273)=300K),(,V_(1)=1.12"litre",V_(2)=?):}` `V_(2)=(760xx1.12)/(273)xx(300)/(760)=1.2308"litre"` |
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| 252. |
If 30mL of `H_(2) and 20mL of `O_(2)`, reacts to form `H_(2)O`, what is left at the end of the reaction?A. 10mL of `H_(2)`B. 5mL of `H_(2)`C. 10mL of `O_(2)`D. 5mL of `O_(2)` |
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Answer» Correct Answer - D `H_(2) (g) +(1)/(2) O_(2)(g) to H_(2)O(g)` `{:(,"Initially", 30, 20,0),(,"After the reaction",0,(20-15),30):}` |
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| 253. |
10mL of N-HCl, 20mL of `N//2 H_(2)SO_(4) and 30mL N//3 HNO_(3)` are mixed togeher and volume made to one litre. The normally of `H^(+)` in the resulting solution is:A. 3N/100B. N/10C. N/20D. N/40 |
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Answer» Correct Answer - A |
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| 254. |
50 " mL of " a gaseous hydrocarbon A requried for complete combustion. 357 " mL of " air `(21%` oxygen by volume) and gaseous products occupied 327 mL (all volumes being measured at STP. The molecular formula of the hydrocarbon A is (a). `C_2H_6` (b). `C_2H_4` (c). `C_3H_6` (d). `C_3H_6` |
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Answer» Let the formula of the hydrocarbon A be CxHy. `C_(x)H_(y)(g)+(x+(y)/(4))O_2(g)toxCO_2(g)+(y)/(2)H_2O(1)` Volume of `O_2=(357xx21)j/(j10j0)=75mL` Volume of `N_2=357-75=282ml` volume of `CO_2=327-282=45mL` `therefore15x=45`, `x=3` `({:(15(x+(y)/(4))=75),(x+(y)/(4)=5):})` `3+(y)/(4)=5` On solving we get `y=8` Formula of A is `C_3H_8` |
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| 255. |
Oxidation numbers of two `Cl` atoms in belaching powder, `CaOCl_(2)`, areA. `+1 only`B. `-1 only`C. `+1 and -1`D. none of these |
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Answer» Correct Answer - C |
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| 256. |
3.0 g sample of KOCl and `CaOCl_(2)` is dissolved in water to prepare 100 mL solution, which requried 100 " mL of " 0.15 M acidified `K_(2)C_(2)O_(4)`. For the point. The clear solution is now treated with excess of `AgNO_(3)` solution which precipitates 2.87 g of `AgCl`. Calculate the mass percentage of KOCl and `CaOCl_(2)` in the mixture. |
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Answer» (a). `2e^(-)+undersetunderset(x=+1)(x-2=-1)(overset(+1)(ClO^(ɵ))toundersetx=-1)(Cl^(ɵ))(n=2)` (b). `2e^(-)+undersetnderset(2x=0)(2x-2=-2)(Cl_(2)O^(-2))tounderset(2x=-2)(Cl^(ɵ))(n=2)` (c). `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)(n=2)` Let a and b millimoles of KOCl and `CaCOCl_(2)` are present in the mixture. `m" Eq of "KOCl+m" Eq of "CaOCl_(2)=m" Eq of "K_(2)C_(2)O_(4)` `2a+2b=100xx0.15xx2` (n-factor) `therefore2a+2b=30` .(i) Also millomoles of `Cl^(ɵ)` from `KOCl+` millimoles of `Cl^(ɵ)` from `CaOCl_(2)-=` millimoles of AgCl` `a(lCl^(ɵ)` ions) `+2b(2Cl^(ɵ)` ions)`=(2.87)/(143.5)xx10^(3)` `[Mw of AgCl=143.5]` `thereforea+2b=20` (ii) From equation (i) and (ii) `a=10,b=5` `% of KOCl=(10xx10^(-3)xx90.5xx100)/(3)` `[Mw of KOCl=90.5]` `=30.1%` `% of CaOCl_(2)=(5xx10^(-3)xx127)/(3)xx100` `[Mw of CaOCl_(2)=127]` `=21.1%` |
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| 257. |
A mixture containing `As_(2)O_(3)` and `As_(2)O_(3)` requried 20 " mL of " 0.05 N `I_(2)` for titration. The resulting solution is then acidified and excess KI was added. The liberated `I_(2)` required 1.24 g hypo `(Na_(2)S_(2)O_(3).H_(2)O)` for complete reaction. Calculate the mass of the mixture. The reactions are `As_(2)O_(3)+2I_(2)+2H_(2)OtoAs_(2)O_(3)+4H^(o+)+4I^(ɵ)` `As_(2)O_(5)+4H^(o+)+4I^(ɵ)toAs_(2)O_(3)+2I_(2)+2H_(2)O` |
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Answer» l" Eq of "`I_(2)` used `=20xx0.05=1.0` Let a and b are m" Eq of "`As_(2)O` and `As_(2)O_(5)` respectively on addition of `I_(2)` to mixture `As_(2)^(3+)` is converted `As_(2)^(5+)`. `thereforem" Eq of "As_(2)O_(3)-=m" Eq of "I_(2) used-=1.0-=m" Eq of "As_(2)^(5)` formed. `thereforea=1.0` After reaction with `I_(2)`, mixture contains all the arsenic in `+5` oxidation state which is then titrated using `KI+` hypo. Thus, `m" Eq of "As_(2)_(3) as As_(2)^(5+) m" Eq of "As(2)O_(5) as As^(5+)=m" Eq of "I_(2)` liberated `=m" Eq of "`hypo used `thereforea+b=(1.24)/(248)xx10^(3)` Ew of `Na_(2)S_(2)O_(5) as 5H_(2)O=(248)/(1)` `thereforea+b=5` ...(ii) weight of `As_(2)O_(3)=1.0xx10^(-3)xx(198)/(4)=0.0495g` `{:(As_(2)^(3+)toAs_(2)^(5+)+4e^(-)),(Ew of As_(2)O_(3)=(198)/(4)):}` Weight of `As_(2)O_(3)=4xx10^(-3)xx(230)/(4)` `(Ew of As_(2)O_(5)=(230)/(4))` `=0.23 g` Weight of mixture`=0.0495+0.23=0.2795g` |
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| 258. |
Hydroxylamine reduces `Fe^(3+)` accoeding to the following reaction: `2NH_2OH+4Fe^(3+)toH_2O+4Fe^(2+)+4H^(o+)+N_2O` `Fe^(2+)` produced is is estimated by titration with `KMnO_4` solution A 10 mL sample of `NH_2OH` is diluted to 1000 mL. 50 " mL of " this diluted sample is boiled with excess of Fe (III) solution. The resulting solution required 12 " mL of " 0.02 M `KMnO_4` for complete oxidation. Determine the strength of `NH_2OH`. |
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Answer» `2NH_(2)OH+4Fe^(3+)toN_(2)O+4Fe^(2+)+H_(2)O+4H^(o+)` `MnO_(4)^(ɵ)+5Fe^(2+)+8H^(o+)toMn^(2+)+4H_(2)O` 12 " mL of " 0.02 of M `KMnO_(4)=12xx0.2xx5mEq` of `Fe(2+)` `=1.2m" " mol of ""Fe^(2+)` `=1.2xx10^(-3)xx56g` of `Fe^(2+)` `4xx56g` of `Fe^(2+)=2xx33g` of `NH_(2)OH` `0.067g` of `Fe^(2+)=(2xx33xx0.672)/(4xx56)` `=0.0198g` of `NH_(2)OH` 0.0198 g of `NH_(2)OH` is present in 50 " mL of " solution `therefore` 1000 " mL of " solution will contain `=(0.0198xx1000)/(50)` `=0.396g` of `NH_(2)OH` This much dilute solution is obtained from 10 " mL of " the original sample. Hence strength `=(0.396xx1000)/(10)=39.6gL^(-1)` Alternatively: `undersetunderset(2x=-2)(2x+6-4=0)(2NH_2OH)toundersetunderset(2x=2)(2x-2=0)(N_2O+4e^(-))(x=(4)/(2)=2)` " Eq of "`NH_2OH=2xx"moles of "NH_2OH(Ew=(33)/(2))` `12mL" of "0.02 M KMnO_4-=12xx0.02xx5mEq" of "Fe^(2+)` `-=1.2mEq" of "NH_2OH` `-=1.2xx10^(-3)xx(33)/(2)g" of "NH_2OH` `-=0.0198 g" of "NH_2OH` After that, follow the above method. |
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| 259. |
`MnO_(2)` on ignition converts into `Mn_(3)O_(4)`. A sample of pyrolusite having 75% `MnO_(2)`, 20% inert impurities and rest water is ignited in air to constant mass. What is the percentage of Mn in the ignited sample ?A. 0.246B. 0.37C. 0.5524D. 0.7405 |
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Answer» Correct Answer - C |
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| 260. |
50.0 g sample of brass is dissolved in 1 L dil `H_2SO_4.20mL` of this solution is mixed with KI, and the liberated `I_2` required 20 " mL of " 0.5 M hypo solution for titration calculate the amount of Cu in the alloy. |
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Answer» `Cu-=CuSO_4-=KI-=I_2-=Na_2S_2O_3` `mEq-=mEq-=mEq-=mEq` `?-=--=--=20xx0.5xx1(n=1)` `=(10 m" Eq of "Na_2S_2O_3)/(20 mL "of solution")` `-=(10xx1000)/(20)mEqL^(-1)=500mEqL^(-1)` of solution m" Eq of "Cu `=500xx10^(-3)xx(63.5)/(1)` (n-factor `=1`) `=(635xx5)/(100)=31.75g` `%` `Cu=(31.75)/(50)xx100=63.5%` |
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| 261. |
32g of a sample of `FeSO_(4) .7H_(2)O` were dissolved in dilute sulphuric aid and water and its volue was made up to 1litre. 25mL of this solution required 20mL of `0.02 M KMnO_(4)` solution for complete oxidation. Calculate the mass% of `FeSO_(4) .7 H_(2)O` in the sample.A. 34.75B. 69.5C. 89.5D. none of these |
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Answer» Correct Answer - A |
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| 262. |
A polyvalent metal weghing 0.1 g and having atomic weight 51 reacted will dil `H_2SO_4` to give 43.9 " mL of " `H_2` at STP. This solution containing the metal in the lower oxidation state was found to require 58.8 " mL of " 0.1 permanganate for complete oxidation. What are the valencies of the metal. |
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Answer» 43.9 " mL of " `H_2` at STP is produced form 0.1 g of metal `11200 " mL of " H_2` at STP produced `=(0.1xx11200)/(43.9)` `=25.5g` of metal Valency of metal is its lower oxidation state. So Valency`=(51)/(25.5)=2` `therefore` Valency`=`Atomic weight/Equivalent weight 0.1 g of the metal in lower oxidation state required 58.8 " mL of " 0.1 `KMnO_4` 51 g of the metal requires `=(58.8xx0.1)/(0.1)xx51` `=2958 m" Eq of "KMnO_4` `=2.9 " Eq of "KMnO_4` `approx3 " Eq of "KMnO_4` Hence, increase in valency during further oxidation is there. Thus higher oxidation state`=5` Valency of metal`=2,5` |
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| 263. |
0.10 g of a sample containing `CuCO_(3)` and some inert impurity was dissolved in dilute sulphuric acid and volume made up to 520 mL. This solution was added into 50mL of 0.04 M KI solution where copper precipitates as Cul and `I^(-)` is oxidized into `I_(3)^(-)`. A 10 mL portion of this solution is taken for analysis, filtered and made up free `I_(3)^(-)` and then treated with excess of acidic permanganate solution. Liberated iodine required 20 mL of 2.5 mM sodium thiosulphate solution to reach the end point. Determine mass percentage of `CuCO_(3)` in the original sample.A. 7.41B. 74.1C. 61.75D. None of these |
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Answer» Correct Answer - B |
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| 264. |
A sample of magnisium metal containing some `MgO` as impurity was dissolved in 125 " mL of " 0.1 N `H_2SO_4`. The volume of `H_2` evolved at `27.3^@C` and 1 atm was 120.1 mL. The resulting solution was found to be 0.02 N with respect to `H_2SO_4`. Calculate (i) the weight of sample dissolved and (ii) the percentage by weight of Mg in the sample. Neglect any change in the volume of the solution (atomic weight of `Mg=24.3`). |
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Answer» Volume of `H_2` at `STP=(120.1xx273)/(300.3)=109.2mL` `Mg+H_2SO_4toMgSO_4+H_2` `MgO+H_2SO_4toMgSO_4+H_2O` Weight of Mg`=(24.3)/(22400)xx109.2=0.1185g` `" Eq of "Mg=(0.1185)/(12.15)=0.009753` `" Eq of "H_2SO_4(t otal)=(125xx0.1)/(1000)=0.0125` Excess of `H_2SO_4=(125xx0.02)/(1000)=0.0025` Excess of `H_2SO_4=(125xx0.2)/(1000)=0.0025` `H_2SO_4` used`=0.0125-0.0025=0.01` `=g" Eq of "Mg+g" Eq of "MgO` " Eq of "MgO`=` Total `Eq-" Eq of "Mg` `=0.01-0.009753` `0.000247 " Eq of "MgO` `=0.00247xx(40.3)/(2)g of MgO` `=0.005g` Total weight `=` Weight of `Mg+` weight of `MgO` `=0.1185+0.005=0.1235g` `% of Mg=(0.1185)/(0.1235)xx100=95.5%` |
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| 265. |
Calculate the quantity of lime required to soften `10^(3)L` of `H_2O` which contains `7.5 g L^(-1)` of `Ca(HCO_3)_2` and `5.0 g L^(-1) of Mg(HCO_3)_2` |
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Answer» (a). `Ca(HCO_(3))_(3)+Ca(OH)_(2)to2CaCO_(3)+2H_(2)O` (b). `Mg(HCO_(3))_(2)+Ca(OH)_(2)toCaCO_(3)+cancel(MgCO_(3))+2H_(2)O` `underline(cancel(MgCO_(3))+Ca(OH)_(2)toMg(OH)_(2)+CaCO_(3))` `underline(Mg(HCO_(3))_(2)+2Ca(OH)_(2)to2CaCO_(3)+Mg(OH)_(2)+2H_(2)O)` `[mW of Ca(HCO_(3))_(3),Mg(HCO_(3))_(2), and Ca(OH)_(2)`, Respectively, are : 162,146 and 74 `mol^(-1)]` From reaction (a), Weight of `Ca(OH)_(2)=(74xx7.5)/(162)=3.42g` From reaction (b), Weight of `Ca(OH)_(2)=(2xx74xx5.0)/(146)=5.06g` Total weight of `Ca(OH)_(2)` in `1 L" of "H_(2)O=3.42+5.06=8.48L^(-1)` Weight of `Ca(OH)_(2)` in `10^(3)L=8.48xx10^(3)g` |
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| 266. |
1L of pond water contains `20mg of Ca^(2+) and 12mg "of" Mg^(2+) ions`. What is the volume of a `2N Na_(2)CO_(3)` solution required to soften 5000L of pond water?A. 500LB. 50LC. 5LD. none of these |
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Answer» Correct Answer - C |
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| 267. |
The hardness of water due to `HCO_3` is `122 p p m`. Select the correct statement(s).A. The hardness of water in terms of `CaCO_(3)` is 200 ppm.B. The hardness of water in terms of `CaCO_(3)` is 100 ppm.C. The hardness of water in terms of `CaCl_(2)` is 222 ppm.D. The hardness of water in terms of `MgCl_(2)` is 95 ppm. |
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Answer» Correct Answer - B::D Since 2 " mol of "`HCO_(3)^(ɵ)` is present. So there should be one mole each of `CaCO_(3)`, `CaCl_(2)` and `MgCl_(2)` to have equal hardness. `thereforeMw of HCO_(3)^(ɵ)=61`, `ppm of HCO_(3)^(ɵ)=61xx2=122 g` in `10^(6) mL H_(2)O` `[Mw of CaCO_(3)=100,Mw of CaCl_(2)=111,Mw of MgCl_(2)=95g mol^(-1)]` `1 " mol of "CaCO_(3)-=100ppm` `1 " mol of "CaCl_(2)=111 ppm` `1 " mol of "MgCl_(2)=95 ppm` Hence, answer is (b) and (d). |
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| 268. |
A certain public water supply contains 0.10ppb (part per billion) of chloroform `(CHCl_(3))`. How many molecules of `CHCl_(3)` would be obtained in 0.478mL drop of this water? (Assuming d=1g//Ml)A. `4xx10^(-13)xxN_(A)`B. `10^(-3)xxN_(A)`C. `4xx10^(-10)xxN_(A)`D. none of these |
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Answer» Correct Answer - A |
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| 269. |
Decreasing order (first having highest and then other following it) of mass of pure NaOH in each of the aqueous solution (P) 50 gm of `40%(w//w)` NaOH (Q) 50 gm of `50%(w//w)` NaOH `[d_("soln.")=1.2gm//ml]` (R) 50 gm of 20 M NaOH `[d_("soln").=1gm//ml]`A. I,ii,iiiB. iii,ii,iC. ii,iii,iD. ii,I,iii |
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Answer» Correct Answer - B |
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| 270. |
A sample of pure CuO was reduced with `H_2` gas and `H_(2)O` formed was collected in a 44.8 L flask containing dry `N_(2)`. At `27^@C`, the total pressure containing `N_(2)` and `H_(2)O` was 1.0 atm. The relative humidity in the flask was `80%` The vapour pressure of water at `27^@C` is 25 mm. How many grams of CuO was reduced? |
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Answer» `CuO+H_(2)toCu+H_(2)O(g)` Volume of dry `N_(2)=44.8 L at 27^@C` Total pressure `(N_(2)+H_(2)O)=1.0atm` Vapour pressure of `H_(2)O=25mm` Humidity `-=80%` Pressure due to `H_(2)O` vapour `=0.8xx(25)/(760)mm` Moles of `H_(2)O` vapour`=(PV)/(RT)=(0.8xx(25)/(760)xx44.8)/(0.082xx300)` According to the above equation: 1 " mol of "`H_(2)O` is obtained from 1 " mol of "CuO. Therefore, moles of CuO reduced`=0.0478` Weight of `CuO=0.0478xx(63.5+16)=3.8g` |
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| 271. |
Calculate the quantities of reagents required to soften `10^(3)` L of water containing `Ca(HCO_3)_2` `Mg(HCO_3)_2` and `CaSO_4` as 20.0 g, 15.0 g and 5.0 g per litre respectively by lime soda process. |
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Answer» `20.0 g of Ca(HCO_(3))_(2)-=(20xx100)/(162)=12.34 g of CaCO_(3)` `15.0 g of Mg(HCO_(3))_(2)-=(15xx100)/(146)=10.27g of CaCO_(3)` `5.0 of CaCO_(4)-=(5xx100)/(136)=3.67g of CaCO_(3)` Lime requirement: `=(74)/(100)` [ Temporary Ca- Hardness`+2xx` Temporary Mg-Hardness `+` Permanent Mg-hardness] `=(74)/(100)[12.34+2xx10.27]` `=(24.33g)/(L)` of `CaCO_(3)eq` `=24.33xx(10^(3)g)/(10^(3)L)` `=(24.33gk)/(10^(3)L) of CaCO^(3) eq` Soda requirement: `=(106)/(100)` [permanent Ca-hardnes`s^(+)` Permanent Mg-hardness] `=(106)/(100)[3.67]` `k=(3.89g)/(L) of CaCO_(3)` equivalent `=3.89xx(10^(3)g)/(10^(3)L)` of `CaCO_(3)eq` `=(3.89kg)/(10^(3)L) of CaCO_(3) eq` |
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| 272. |
Permanent hardness is due to `CI^(ɵ)` and `SO_4^(2-)` of `Mg^(2+)` and `Ca^(2+)` and is removed by adding `Na_2CO_3`. `{:(CaSO_(4)+Na_(2)CO_(3)toCaCO_(3)+Na_(2)SO_(4)),(CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl):}` Which of the following statements is`//`are correct?A. If hardness is 100 ppm `CaCO_(3)` the amount of `Na_(2)CO_(3)` requried to soften 10 L of hard water is 10.6 g.B. If hardness is 100 ppm `CaCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard is 10.6 gC. If hardness is 420 ppm `MgCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard water is 53.0 g.D. If hardness is 420 ppm `MgCO_(3)` the amount of `Na_(2)CO_(3)` required to soften 10 L of hard water is 5.3 g. |
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Answer» Correct Answer - A::D Mw of `CaCO_(3)+40+12+3xx16=100gmol^(-1)` Mw of `Na_(2)CO_(3)=46+60=106 g mol^(-1)` (i). `underset(x mol)(CaSO_(4))+underset(x mol)(Na_(2)CO_(3))toCaCO_(3)+Na_(2)SO_(4)` ..(i) (ii). `underset(y mol)(CaCl_(2))+underset(y mol)(Na_(2)CO_(3))toCaCO_(3)+2NaCl` .(ii) (iii). 100 ppm `CaCO_(3)-=100g CaCO_(3)` in `10^(6)mL` `=(100)/(100)mol` in `10^(6)mL` Moles of `Na_(2)CO_(3)` required`=" mol of "CaCO_(3)` `-=(100)/(100)mol` in `10^(6)mL` `-=1 m" mol of "10^(6)mL` `-=(1)/(10^(6))xx10xx10^(3)mL(10L)` `=1xx10^(-2) mol` in `10L` Therefore, moles of `Na_(2)CO_(3)` required `=1xx10^(-2) mol` in `10L` `=1xx10^(-2)xx(106g)/(10L)` `=1.06g` Similarly for 100 ppm `MgCO_(3):(Mw of MgCO_(3)=24+60=84gmol^(-1))` (iv). `underset(xmol)(MgSO_(4))+underset(xmol)(Na_(2)CO_(3))toMgCO_(3)+Na_(2)SO_(4)`. (i) (v). `underset(y mol)(MgCl_(2))+underset(y mol)(Na_(2)CO_(3))toMgCO_(3)+2NaCl` ..(ii) (vi). `420 ppm MgCO_(3)-=420g MgCO_(3)` in `10^(6)` mL `-=(420)/(84)=5 mol` in `10^(6)mL` Moles of `Na_(2)CO_(3)` required `-=" mol of "MgCO_(3)` `-=5 mol` in `10^(6)mL` `-=(5)/(10^(6))xx10xx10^(3)mL(10L)` `-=5xx10^(-2) mol in 10L` Moles of `Na_(2)CO_(3)` required `-=5xx10^(-2)mol(10L)^(-1)` `-=5xx10^(-2)xx106g(10L)^(-1)` `=(5.3g)/(10L)` |
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| 273. |
`2 "mole" N_(2)` and `3 "mole" H_(2)` gas are allowed to react in a `20 L` flask at `400 K` and after complete conversion of `H_(2)` into `NH_(3)`. `10 L H_(2)O` was added and temperature reduced to `300 K`. Pressure of the gas after reaction is : `N_(2)+3H_(2)to2NH_(3)`A. `3Rxx(300)/(20)`B. `3Rxx(300)/(10)`C. `Rxx(300)/(20)`D. `Rxx(300)/(10)` |
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Answer» Correct Answer - D `N_(2)` left `=1 mol` `NH_(3)` formed`=2 mol` but dissolved in `H_(2)O`, hence pressure is due to `N_(2)` only Hence, volume of the flask is `10 L` since 10 L `H_(2)O` is added. `thereforeP=(n)/(V)RT=(300R)/(10)` |
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| 274. |
If water contains 10 ppm of `MgCl_2` and 8 ppm of `CaSO_4`, calculate the ppm of `CaCO_3`. |
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Answer» `MgCl_(2)=(10xx100)/(65)=10.53ppm`, `CaSO_(4)=(8xx100)/(136)=5.92ppm` `ppm of CaCO_(3)=10.53+5.92=16.45ppm`. |
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| 275. |
A 200 mL sample of water requires 5 " mL of " `(N)/(20)Na_2CO_3` solution for complete precipitation of `Ca^(2+)` and `CaCO_3`. Calculate the hardness in ppm. |
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Answer» `5 " mL of " (N)/(20) Na_(2)CO_(3)=(5xx50)/(20xx1000)g CaCO_(3)` So, 200 parts of water contains`=(5xx50)/(20xx1000)` parts of `CaCO_(3)` equivalent `10^(6)` parts of water contains`=(5xx50xx10^(6))/(20xx1000xx200)` `=62.5 ppm` Therefore, hardness`=62.5 ppm` |
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| 276. |
Hardness of water is 200 ppm. The normality and molarity of `CaCO_3` in the water is (a). `2xx10^(-6)(N,2xx10^(-6)M` (b). `4xx10^(-2)N,2xx10^(-2)M` (c). `4xx10^(-3)N,2xx10^(-3)M` (d). `4xx10^(-1)N,2xx10^(-1)M` |
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Answer» Hardness is weight in grams of `CaCO_3` in `10^(6) mL` of `H_2O` `or=10^(6)` of `H_2O` `(becausedH_2O=1)` `Mw(CaCO_3)=100gmol^(-1)` `M=(W_2xx1000)/(Mwxx"Volume of solutio in mL")` `=(200xx1000)/(100xx10^(6))=2xx10^(-3)M` `N=nxxM=2xx2xx10^(-3)=4xx10^(-3)N` |
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| 277. |
For the redox reation `MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)CO_(2)+H_(2)O` The correct stoichiometric coefficients of `Mno_(4)^(-),C_(2)O_(4)^(2-)` and `H^(+)` respectively:A. 2,5,16B. 16,3,12C. 15,16,12D. 2,16,5 |
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Answer» Correct Answer - A |
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| 278. |
Three different solution of oxidising agents. `K_(2)Cr_(2)O_(7),I_(2), and KMnO_(4)` is titrated separately with `0.19g of K_(2)S_(2)O_(3)`. The molarity of each oxidising agent is 0.1 M and the reaction are: (i). `Cr_(2)O_(7)^(2-)+S_(2)O_(3)^(2-)toCr^(3+)+SO_(4)^(2-)` (ii). `I_(2)+S_(2)O_(3)^(2-)toI^(ɵ)+S_(4)O_(6)^(2-)` (iii). `MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)toMnO_(2)+SO_(4)^(2-)` (Molecular weight of `K_(2)S_(2)O_(3)=190,K_(2)Cr_(2)O_(7)=294,KMnO_(4)=158, and I_(2)=254 mol^(-1))` Which of the following statements is/are correct?A. All three oxidising agents can act as self-indicatorsB. Volume of `I_(2)` used in minimum.C. Volume of `K_(2)Cr_(2)O_(7)` used in maximum.D. weight of `KMnO_(4)` used in the titration is maximum. |
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Answer» Correct Answer - A::B::D All three are self indicato rs i.e., they do not need any indicator for titration. `K_(2)S_(2)O_(3)=(0.19)/(190)=10^(-3)mol=1mmol` (i). `m" Eq of "S_(2)O_(3)^(2-)(n=-8)=m" Eq of "Cr_(2)O_(7)^(2-)(n=6)` `1xx8=0.1xx6xxV` `V_(Cr_(2)O_(7)^(2-))=(80)/(6)mL` (ii). `m" Eq of "S_(2)O_(3)^(2-)(n=1)=m" Eq of "I_(2)(n=2)` `1xx1=0.1xx2xxV` `V_(I_(2))=5mL` (iii). `m" Eq of "S_(2)O_(3)^(2-)(n=8)=m" Eq of "MnO_(4)^(ɵ)(n=3)` `1xx8=0.1xx3xxV` `thereforeV_(MnO_(4)^(ɵ))=(80)/(3)mL` Also, Weight of `K_(2)Cr_(2)O^(7)=8xx(294)/(6)xx10^(-3)=0.392g` Weight of `I_(2)=1xx(254)/(2)xx10^(-3)=0.127g` Weight of `KMnO_(4)=8xx(158)/(3)xx10^(-3)=1.274g` |
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| 279. |
How many " mL of " 0.1 N HCl is required to react completely using phenolphthalein with 2.0 g mixture of `Na_2CO_3` and `NaOH` containing equimolar amounts of two? |
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Answer» Let x mol each of `Na_2CO_3` and `NaOH` be there in `2.0g` Mixture. So `x(106)+x(40)=2impliesx=(1)/(73)` In neutralisation: g " Eq of "`Na_2CO_3=2x` But only half will be neutralised in phenolpthalein. g " Eq of "`NaOH=x`, fully neutralised `implies(1)/(2)(2x)+x=g m" Eq of "HCl=(1)/(10)V` `implies2x=(V)/(10)` or `V=20x=20((1)/(73))=0.274L-=274mL` |
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| 280. |
20 " mL of " x M HCl neutralises completely 10 " mL of " 0.1 M `NaHCO_3` solution and a further 5 " mL of " 0.2 M `Na_2CO_3` to methyl orange end point. What is the value of `x`? |
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Answer» `mEq` of `HCl=mEq` of `NaHCO_3+mEq` of `Na_2CO_3` `20x xx=10xx0.1+5xx0.2xx2` `x=(3)/(20)=0.15M` |
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| 281. |
20 " mL of " x M HCl neutralises completely 10 " mL of " 0.1 M `NaHCO_3` solution and a further 5 " mL of " 0.2 M `Na_2CO_3` to methyl orange end point. What is the value of `x`?A. 0.167 MB. 0.133 MC. 0.15 MD. 0.2 M |
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Answer» Correct Answer - C 20 " mL of " x M HCl`-=10" mL of " 0.1 M `NaHCO_(3)+5" mL of " 0.2 M Na_(2)CO_(3)` (methyl orage indicated `100%` reaction of `NaHCO_(4) and Na_(2)CO_(3)`) 0.2 M `Na_(2)CO_(3)=0.4Na_(2)CO_(3)` `20x=1+2impliesx=0.15` |
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| 282. |
200 " mL of " a solution of a mixture of NaOH and `Na_2CO_3` was first titrated with 0.1 M HCl using phenolphthalein indicator. 17.5 " mL of " HCl was required for the same HCl was again required for next end point. Find the amount of NaOH and `Na_2CO_3` in the mixture. |
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Answer» With phenolphthalein `m" Eq of "Hcl` used for 200 mL solution`=17.5xx0.1xx1` `=1.75` Let a and b m" Eq of "NaOH and `Na_(2)CO_(3)` respectively. `a+(b)/(2)=1.75` .(i) With methyl orange indicator m" Eq of "HCl used for solution after first end point `=2.5xx0.1xx1=0.25` `(b)/(2)=0.25` .(ii) By equations (i) and (ii) `a(mEq. of NaOH)=1.5` Weight of NaOH`=1.5xx10^(-3)xx40` `=(0.06g)/(200mL)` `b(m" Eq of "Na_(2)CO_(3))=0.5` Weight of `Na_(2)CO_(3)=0.5xx10^(-3)xx(106)/(2)` `=(0.0265)/(200mL)` |
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| 283. |
30 " mL of " a solution of mixture of `Na_2CO_3` and `NaHCO_3` required 12 " mL of " 0.05 M `H_2SO_4` using phenolphthalein as indicator. With methyl orange 30 " mL of " the same solution required 40 " mL of " same `H_2SO_4`. Calculate the amount of `Na_2CO_3` and `NaHCO_3` per litre in the mnixture. |
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Answer» With phenolphthalein indicator `m" Eq of "H_(2)SO_(4) used =12xx0.05xx2` (n-factor) `=1.2` Let a and b m" Eq of "`NaHCO_(3)` and `Na_(2)CO_(3)`, respectively. `therefore(b)/(2)=1.2` With methyl orange indicator `m" Eq of "H_(2)SO_(4)` used from the begining `=4.0xx0.05xx2` `=4.0` `a+b=4.0` By equations (i) and (ii) we get `a(m" Eq of "NaHCO_(3))=1.6` Weight of `NaHCO_(3)=1.6xx10^(-3)xx84` `=(0.1344g)/(30mL)` `=(0.1344xx1000)/(30)gL^(-1)` `=4.48 g L^(-1)` `b(m" Eq of "Na_(2)CO_(3))=2.4` Weight of `Na_(2)CO_(3)=2.4xx10^(-3)xx(106)/(2)` `=(0.1272g)/(30mL)` `=(0.1272xx1000)/(30)gL^(-1)` `=4.24 g L^(-1)` |
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| 284. |
In the following equation `CrO_(4)^(2-)+S_(2)O_(3)^(2-)+overset(ɵ)(O)Hto[Cr(OH)_(4)]^(-1)+SO_(4)^(2-)` What volume of 0.2 M `Na_(2)S_(2)O_(3)` solution.A. 30 mLB. 80 mLC. 20 mLD. 60mL |
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Answer» Correct Answer - B (i). `3e^(-)+undersetunderset(x=6)(x-8=-2)(CrO_(4)^(2-))toundersetunderset(2x=12)(2x-6=-4)([Cr(OH)_(4)]^(ɵ))` Ew of ``CrO_(4)^(2-)=(Mw)/(3)` (ii). `undersetunderset(2x=4)(2x-6=-2)(S_(2)O_(3)^(2-)toundersetunderset(2x=12)(2x-6=-4)(2SO_(4)^(2-)+8e^(-)` " Eq of "`S_(2)O_(3)^(2-)=(Mw)/(8)` `N_(1)V_(1)(CrO_(4)^(2-))=N_(2)V_(2)(S_(2)O_(3)^(2-))` i.e., `m" Eq of "CrO_(4)^(2-)=m" Eq of "S_(2)O_(3)^(2-)` `V_(1)xx0.2xx3=30xx0.2xx8` `V_(1)=(30xx0.2xx8)/(0.2xx3)=80mL` |
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| 285. |
80 " mL of " `KMnO_4` solution reacts with 3.4 g of `Na_2C_2O_4.2H_2O` in acidic medium. The molarity of the `KMnO_4` solution is. (a). 0.5M (b). 0.1M (c). 5M (d). 1M |
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Answer» (b). `Mw[Na_2C_2O_4.2H_2O]` `=2xx23+2xx12+4xx16+2xx18=170g` `[5e^(-)+MnO_4^(ɵ)toMn^(2+)](n=5)` `[C_2O_4^(2-)to2CO_2+2e^(-)](n=2)` `MnO_4^(ɵ)-=C_2O_4^(2-)` `1mEq-=1mEq` `80xxN-=(wxx10^(3))/(Ew)mEq=(3.4xx10^(3))/((170)/(2))` `N(MnO_4^(ɵ))-=(3.4xx10^(3))/((170)/(2)xx80)=0.5N` `M(MnO_4^(ɵ))-=(0.5N)/(5)=0.1`M |
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| 286. |
A mixutre solution of KOH and `Na_2CO_3` requires 15 " mL of " `(N)/(20)` HCl when titrated with phenolphthalein as indicator.But the same amoound of the solutions when titrated with methyl orange as indicator requires 25 " mL of " the same acid. Calculate the amount of KOH and `Na_2CO_3` present in the solution. |
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Answer» (a). With methyl orange as indicator: Full titre value of KOH (let x mL)`+` full titre value of `Na_2CO_3` (let y mL)`=25mL` `x+y=25` .(i) (b). With phenolphthalein as indicator: Full titre value of KOH(x)`+(1)/(2)` titre value of `Na_2CO_3` `((y)/(2))=15mL` `x+(y)/(2)=15` ...(ii) Solving (i) and (ii) we get `x=5mL, y=20mL` Full titre value of `Na_2CO_3=20mL` m" Eq of "`Na_2CO_3=mEq ` of HCl `=20xx(N)/(20)mEq` of `Na_2CO_3` `=1xx10^(-3)xx53g of Na_2CO_3` `=0.053g` Volume of KOH`=25-20=5mEq` of `KOH` `=5xx10^(-3)xx56g` of `KOH` `=0.014` g of `KOH` |
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| 287. |
0.4 g of a mixutre containing sodium oxalate `(Na_2C_2O_4)` and potassium oxalate requires 50 " mL of " `(M)/(60)K_2Cr_2O_7` solution in acidic medium for complete reaction. Calculate the percentage composition of the mixture. |
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Answer» `Ew(Na_2C_2O_4)=(2xx23+2xx12+4xx16)/(2)=(134)/(2)=67g` `Ew(K_2C_2O_4)=(2xx39+2xx12+4xx16)/(2)=(166)/(2)=83g` Let x g of `Na_2C_2O_4` and `(0.4-x)g of K_2Cr_2O_7` `((x)/(67)+(0.4-x)/(83))xx10^(3)=50xx(1)/(60)xx6` (n-factor) `83x+67(0.4-x)=(5xx67xx83)/(10^(3))` `16x=27.8-67xx0.4` `=27.8-26.8=1.0` `x=(1)/(16)=0.06g` `% of Na_2C_2O_4=(0.06xx100)/(0.4)=15.6%` `% of K_2C_2O_4=100-15.6=84.4%` |
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| 288. |
A sample of `Fe(SO_4)_3` and `FeC_2O_4` was dissolved in `H_2SO_4`. 40 " mL of " `(N)/(16)KMnO_4` was required for complete oxidation. After oxidation the mixture was reduced by `(An)/(H_2SO_4)`. On again oxidation by same `KMnO_4`, 60 mL was required. Calculate the ratio of m" Eq of "`Fe_2(SO_4)_3` and `FeC_2O_4`. |
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Answer» `FeC_2O_4+H_2SO_4toFeSO_4+(COOH)_2` `Fe^(2+)toFe^(3+)+e^(-)` `C_2O_4^(2-)to2CO_2+2e^(-)` 40 " mL of " `(N)/(16)` `KMnO_4=(40)/(16)` m" Eq of "`(Fe^(+2)+OX^(2-))` Let us take y m " mol of "`Fe^(2+)` `therefore m" Eq of "Fe^(+2)=y` (n factor`=1`) y m " mol of "`OX^(2-)` `thereforem" Eq of "OX^(2-)=2y` (n factor`=2`) `thereforey+2y=(40)/(16)implies3y=(40)/(16)becausey=(40)/(48)` m" Eq of "`Fe^(2+)=(40)/(48)`, `m" Eq of "Fe_2C_2O_4=(40)/(16)` Second step: 60 " mL of " `(N)/(16)KMnO_4=(60)/(16)` m" Eq of "total `Fe^(2+)` ltBrgt `[Fe^(2+)` ions from `FeC_2O_4+Fe^(2+)` ions obtained after the reduction of `Fe_2(SO_4)_3]` m" Eq of "`Fe_2(SO_4)_3=(60)/(16)-(48)/(48)=(140)/(48)` Ratio of mEq`(Fe_2(SO_4)_3)/(FeC_2O_4)=(140xx48)/(48xx40)=(7)/(6)=7:6` |
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| 289. |
Calculate the mass of oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39,O=16 and Cl=35.5)A. 39.2kgB. 392kgC. 3.92kgD. 3kg |
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Answer» Correct Answer - C The reaction is : `underset(2xx122.5 kg)(2KClO_(3))(s) to 2KCl(s) + underset(3xx32 kg)(3O_(2)(g))` Mass of `O_(2)` obtained by 10 kg `KClO_(3) = (96 xx 10)/(245) = 3.92 kg `] |
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| 290. |
The molecular formula of a commercial resin used for exchanging ions in water softening is `C_(8)H_(7)SO_(3)Na` (Mol.wt.206). What would be the maximum uptake of `Ca^(2+)` ions by the resin when expressed in mole per gram resin?A. `(1)/(412)`B. `(1)/(103)`C. `(1)/(206)`D. `(2)/(309)` |
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Answer» Correct Answer - A `underset(2xx 206 g)underset("OR")underset("2 mol")(2C_(8) H_(7)SO_(3)Na)+underset("1mol")(Ca^(2+)) to (C_(8)H_(7)SO_(3)) Ca+ 2Na^(+)` Number of moles of `Ca^(2+)` removed by one gram resin `=(1)/(412)` mol] |
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| 291. |
A sample of peanut oil weighing `1.5763 g` is added to `25 mL` of `0.4210 M KOH`. After saponification is complete `8.46 mL` of `0.2732 M H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:A. 146.72B. 223.44C. 98.44D. 98.9 |
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Answer» Correct Answer - A |
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| 292. |
When 100mL solution of `NaOH and NaCO_(3)` was first titrated with N/10 HCl in presence of HPh, 17.5mL were usedtill end point is obtained. After this end point MeOH was added and 2.5mL of same HCl were required to attain new end point. The amount NaOH in mixture is:A. 0.06g per 100mLB. 0.06g per 200mLC. 0.05 g per 100mLD. 0.012 g per 200mL |
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Answer» Correct Answer - A |
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| 293. |
Iodobenzene is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown here: `C_(6)H_(5)NH_(2)+HNO_(2)+HCl to C_(6)H_(5)N_(2)^(+)Cl+2H_(2)O` `C_(6)H_(5)N_(2)^(+)Cl^(-)+KI to C_(6)H_(5)I+N_(2)+KCl` In an actual preparation, 9.30g of aniline was converted to 12.32g of iodobenzene. The percentage yield of iodobenzene is:A. 0.08B. 0.5C. 0.75D. 0.8 |
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Answer» Correct Answer - D `1"mole of "C_(6)H_(5)NH_(2)(123g)-=1"mole of "C_(6)H_(5)I(204g)` `therefore "9.3g of aniline will give"=((204)/(123)xx9.3)"g iodobenzene"` =15.24g iodobenzene `"% yield"=("Actual amount of product")/("Calculated amount of product")xx100` `=(12.32)/(15.424)xx100=80%` |
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| 294. |
10 litre of `O_(2)` gas is reacted with 30 litre of CO gas at STP. Then volumes of each gas present at the end of reaction areLA. `CO("10 litre"),CO_(2)("20 litre")`B. `O_(2)("10 litre"),CO("30 litre")`C. `CO("20 litre"),CO_(2)("10 litre")`D. `O_(2)("10 litre"),CO_(2)("20 litre")` |
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Answer» Correct Answer - A `CO(g)+(1)/(2)O_(2)(g) to CO_(2)(g)` `{:(,t=0,30L,10L,0),(,"After reaction",(30-20)L,0,20L):}` |
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| 295. |
Benzamide can be preapred by the action of concentrated ammonia upon benzoyl chloride. `underset("Benzoyl chloride")(C_(6)H_(5)COl)+2NH_(3) to underset("Benzamide")(C_(6)H_(5)_CONH_(2))+NH_(4)Cl` In such experiment 65cc of concentrated ammonia (in excess) was treated with 15g of benzoyl chloride to give 11.1g of pure benzamide. Molar masses: benzoyl chloride (141), (benzamide)(121). The percentage yield of benzamide is:A. `(11.1)/(15)xx100`B. `((15-11.1))/(15xx100)`C. `(11.1)/(65)xx100`D. `(121)/(141)xx100` |
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Answer» Actual amount of benzamide formed=11.1g `"% yield"=("Actual amount of product")/("Calculated amount of product")xx100` `=(11.1xx100)/(((121)/(141)xx15))=(11.1xx141)/(121xx15)xx100` |
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| 296. |
At T K, 100 litre of dry oxygen is present in a sealed container. It is subjected to silent electric discharge, till the volumes oxygen and ozone become equal. What is the volume (in litre) of ozone formed at T K?A. 50LB. 60LC. 30LD. 40L |
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Answer» Correct Answer - D ` 3O_(2) to 2O_(3)` `t = 0" " 100` After the reaction `100-3x " " 2x` `because " " 100- 3x = 2x or x =20` `therefore " " "Volume of ozone " = 2x = 40 L]` |
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| 297. |
Crude calcium carbide is made in an electric furance by the following reactions: `CaO+3CtoCaC_2+CO`. The product contains `85%` of `CaC_2` and `15%` unreacted CaO. (a). How much CaO is to be added to the fur |
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Answer» (a). `CaO+3CtoCaC_2+CO` All the CaO is not used up. Calculate how much CaO is required for 85 g of `CaC_2`. That much CaO plus 15 g excess should be taken. 15 g remains as it is. `64 g of CaC_2-=56g of CaO` `therefore85g of CaC_2-=56g of CaC` `therefore85g of CaC_2-=56xx(85)/(64)=74.374 g of CaO` Total `CaO-=74.347+15=89.374g` For each 85 g of pure `CaC_2` and 100 g of impure `CaC_2`. 85 kg pure `CaC_2=89.374kg of CaO` `1000kg pure CaCl_2=(89.374xx1000)/(85)` `=1051.46kg of CaO` (b). `100 kg` crude `CaC_2=89.374kg CaO` `therefore1000kg` crude `CaC_2=893.74kg CaO` |
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| 298. |
1.5g `CdCl_(2)` was formed to contain 0.9g Cd. Calculate MO, the atomic weight of Cd.A. 118B. 112C. 106.5D. 53.25 |
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Answer» Correct Answer - C `because` 0.6 g chlorine combines with 0.9 g Cd. `because` Atomic weight of Cd = 105.5] |
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| 299. |
Carnallite is a double chloride of K and Mg containing `38.86%` of water. 0.458 g of it gave 0.71 g AgCl and 0.666 g of it gave 0.27 g magnesium pyrophosphate `(Mg_2P_2O_7)`. Calculate the percentage of KCl in the carnallite. |
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Answer» Only `MgCl_2` can give `Mg_2P_2O_7`. Thus, the percentage of `MgCl_2` can be determined. `2MgCl_2toMg_2P_2O_7` `0.27g of Mg_2P_2O_7-=(190xx0.27)/(222)=0.231 g of MgCl_2` `% of MgCl_2=(0.231xx100)/(0.666)-=34.68%` `% of HCl-=100-(34.68+38.86)-=26.46%` |
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| 300. |
`Al(SO)_(4))_(3)` solution of 1 molal concentration is present in 1 litre solution of density 2.684 g//cc. How many moles `BaSO_(4)` would be precipated on adding excess `BaCl_(2)` in it?A. 2 molesB. 3 molesC. 6 molesD. 12 moles |
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Answer» Correct Answer - C |
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