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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
In the mixture of `(NaHCO_(3)+Na_(2)CO_(3))` volume of HCl required is x mL with phenolphthalein indicator and then y mL with methyl orange indicator in same titration Hence, volume of HCl for complete reaction of `Na_(2)CO_(3)` isA. 2xB. yC. x/2D. (y-x) |
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Answer» Correct Answer - D |
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| 102. |
In the mixture of `(NaHCO_(3)+Na_(2)CO_(3))` volume of HCl required is x mL with phenolphthalein indicator and then y mL with methyl orange indicator in same titration Hence, volume of HCl for complete reaction of `Na_(2)CO_(3)` isA. `2x`B. `y`C. `(x)/(2)`D. `(y-x)` |
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Answer» Correct Answer - D With phenolphthalein indicator: `NaHCO_(3)` does not react with HCl whereas `Na_(2)CO_(#)` reacts upto `NaHCO_(3)` stage `(50%` reaction). `V_(HCl)=x mL` (ii). With methyl orange indicator : `NaHCO_(3)` reacts completely with HCl and with `Na_(2)CO_(3)` is `100%` reaction. But y " mL of " HCl is added after `Na_(2)CO_(3)` has reacted upto `NaHCO_(3)`. (i.e., half titre value of `Na_(2)CO_(3)`) `V_(HCl)=` Full titre value of `NaHCO_(3)+` half titre value of `Na_(2)CO_(3)`. `ymL=` full titre value of `NaHCO_(3)+x mL` Full titre value of `NaHCO_(3)=(y-x)mL` |
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| 103. |
The volatile chloride of an element has a vapour density `approx69`. One gram of the chloride on hydrolysis yields hydrochloric acid and compound free of chlorine. Addition hydrochloric acid and compound free of chlorine. Addition of `AgNO_(3)` to this solution precipitates 3.129 g of AgCl. What may be the atomic weight of the element.? |
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Answer» Molecular weight`=2xx69=138`. Since 1 g of thwe chloride gives 3.129 g of AgCl, it follows that 1 mol, i.e., 138 g, would yield `(138xx3.129)/(143.5)` moles of `AgClapprox3` " mol of "AgCl. Formula of the chloride `=XCl_(3)` Therefore, atomic weight of `X=(138-3xx35.5)=31.5` |
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| 104. |
1 g molecule of `V_(2)O_(5)` contains :A. 5 mole of oxygen atomB. 2 mole of V atomC. 1 mole of oxygen atomD. 2.5 mole of oxygen atom |
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Answer» Correct Answer - A::B |
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| 105. |
Which of the following concentration terms is/are affected by a change in temperature ?A. MolarityB. MolalityC. NormalityD. Specific gravity |
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Answer» Correct Answer - A::C::D |
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| 106. |
Select the dimensionless quantity (ies) :A. vapour densityB. molalityC. specific gravityD. mass fraction |
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Answer» Correct Answer - A::C::D |
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| 107. |
STATEMENTS-1 : Specific gravity is dimensionless. STATEMENTS-2 : Specific gravity is density of a substance measured w.r.t. density of water at `4^(@)C`.A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT-1B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT- is 1 TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - A |
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| 108. |
Which is/are INCORRECT statement ?A. Equivalent mass of `H_(2)PO_(3)^(-)` is 40.5.B. Eq. mass of `H_(2)PO_(4)^(-)` may be equal to molar mass or less than molar mass because it depends on the reaction.C. `KMnO_(4)` has maximum eq. mass in acidic medium.D. Oxidation state of H in `MgH_(2)` is greater than in `H_(2)O_(2)`. |
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Answer» Correct Answer - A::C::D |
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| 109. |
0.1 " mol of "`MnO_(4)^(ɵ)` (in acidic medium) can:A. Oxidise 0.5 " mol of "`Fe^(2+)`B. Oxidise 0.166 " mol of "`FeC_(2)O_(4)`C. Oxidise 0.25 " mol of "`C_(2)O_(4)^(2-)`D. Oxidise 0.6 " mol of "`Cr_(2)O_(7)^(2-)` |
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Answer» Correct Answer - A::B::C (a). `MnO_(4)^(ɵ)+8H^(o+)+5e^(-)toMn^(2-)+4H_(2)O` `m" Eq of "MnO_(4)^(ɵ)=0.1xx5=0.5` `Fe^(2+)toFe^(3+)+e^(-)` `impliesm" Eq of "Fe^(2+)=0.5` (b). `FeC_(2)O_(4)toFe^(3+)+2CO_(2)+3e^(-)` `implies m" Eq of "FeC_(2)O_(4)=0.166xx3=0.5` (d). `Cr_(2)O_(7)^(2-)` being an oxidising agent cannot be oxidised |
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| 110. |
The equivalent mass of HCl in the given reaction is:A. 16.25B. 36.5C. 73D. 85.1 |
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Answer» Correct Answer - D |
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| 111. |
Equivalent weight of `FeS_(2)` in the half reaction `FeS rarr Fe_(2)O_(3) + SO_(2)` is :A. `(M)/(10)`B. `(M)/(11)`C. `(M)/(6)`D. `(M)/(1)` |
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Answer» Correct Answer - B |
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| 112. |
Which of the following does not represent redox reaction?A. `Cr_(2)O_(7)^(2-)+2overset(ɵ)(O)HtoCrO_(4)^(2-)+H_(2)O`B. `SO_(5)^(2-)+2I^(ɵ)+2H^(o+)toI_(2)+SO_(4)^(2-)`C. `2Ca(OH)_(2)+2Cl_(2)toCa(ClO)_(2)+CaCl_(2)+2H_(2)O`D. `PCl_(5)toPCl_(3)+Cl_(2)` |
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Answer» Correct Answer - A In basic medium `Cr_(2)O_(7)^(2-)` changes to `CrO_(4)^(2-)` (no change in oxidation number). |
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| 113. |
50 " mL of " water on titration with standard soap solution gave the following results: Lather factor `=0.4` mL, total hardness `(TH)=8.2mL`, permanent hardness (PH)`=2.5mL` and standard hard water (containing `0.2 g CaCO_3L^(-1))=19.9mL`. Calculate each type of hardness in ppm. |
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Answer» Strength of `CaCO_3` in hard water `=(0.2xx50)/(1000)g L^(-1)[Ew(CaCO_3)=50]` `=0.01g=10mg of CaCO_3 Eq` Deducting lather factor, `TH=8.2-0.4=7.8mL` `PH=2.5-0.4=2.1mL` Volume of soap solution corresponding to 50 " mL of " standard hard water `=19.9-0.4=19.5mL` `19.5` " mL of " soap solution`-=10CaCO_3Eq` 1 " mL of " soap solution `=(10)/(19.5)` `=0.5128 mg CaCO_3` Eq Volume of soap solution for total hardness `=7.8mL` `=7.8xx0.5128mgCaCO_3` equivalents `=(7.8xx0.5128xx10^(3)mL)/(50mL)` `=80mg L^(-1)=80ppm (80g 10^(6)mL)` `TH=80ppm` Volume of soap solution for PH `=2.1mL` `=2.1xx0.5128mg` `=(2.1xx0.5128xx10^(3)mL)/(50mL)` `=21.53mgL^(-1)=21.53(g)/(10^(6))mL` `=21.53ppm` Hence, `PH=21.53ppm` `TH=80.21.53=58.47ppm` |
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| 114. |
A 200 mL sample of hard water requires 33.0 " mL of " 0.01 M `H_2SO_4` for complete neutralisation. 200 " mL of " the same sample was boiled with 15.0 " mL of " 0.1 M NaOH solution, filtered and made up to 200 mL again. This sample now requires 53.6 " mL of " 0.01 M `H_2SO_4`. Calculate Mg hardness. |
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Answer» 33.0 " mL of " `0.01xx2NH_2SO_4` `-=(333.0xx0.01xx2xx50)/(1000)` `-=0.033gCaCO_3(EwCaCO_3=50)` `-=(0.033xx10^(6))/(200mL)-=165ppm of CaCO_3` Eq Thus temporary hardness `-=165ppm` `mEqNaOH=15xx0.1xx1=1.5` `m" Eq of "H_2SO_4=53.6xx0.01xx2=1.072` `m" Eq of "NaOH` (excess) that reacted with water `=1.5-1.072=0.428mEq` `0.428mEq NaOH-=(0.428xx50xx10^(6))/(1000xx200)ppmCaCO_3` `-=107ppm of CaCO_3` Thus Mg hardness `-=107ppm` |
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| 115. |
How much volume of CO2 at STP is liberated by combustion of 100 cm3 of propane? |
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Answer» Balanced equation of combustion reaction |
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| 116. |
A 150 mL of solution of `I_(2)` is divided into two unequal parts. I part reacts with hypo solution solution in acidic medium. 15 mL of 0.4 M hypo was consumed. II part was added with 100 mL of 0.3 MNaOH solution. What was the initial concentration of `I_(2)` ?A. 0.08 MB. 0.1 MC. 0.2 MD. None of these |
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Answer» Correct Answer - B |
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| 117. |
What volume of 0.2 M `KMnO_(4)` is required to react with 1.58 g of hypo solution `(Na_(2)S_(2)O_(3))` in acidc medium?A. 20 mLB. 10 mLC. 16.6 mLD. 50 mL |
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Answer» Correct Answer - B Ew of `Na_(2)S_(2)O_(3)=(158)/(n)(n=1)` `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)(n=(2)/(2)=1)` ltBrgt `MnO_(4)^(ɵ)=S_(2)O_(3)^(2-)` `mEw-=mEq` `0.2 Mxx5` ("n-factor") `xxV=(1.58)/(158)xx10^(3)` `V=10mL` |
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| 118. |
Consider the following series of reactions : `Cl_(2)+2NaOH to NaCl+NaClO+H_(2)O` `3NaClO to 2NaCl+NaClO_(3)` `4NaClO_(3) to 3NaClO_(4)+NaCl` How many moles of NaCl will be formed by using 1 mole `Cl_(2)` and other reagents in excess ?A. `(1)/(12)` moleB. 1.67 moleC. 1.75 moleD. 0.75 mole |
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Answer» Correct Answer - C |
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| 119. |
If hardness of water sample is 200ppm, then select the incorrect statement:A. `"Mass ratio of"CaCO_(3) "to" H_(2)O is (0.02)/(100)`B. `"Mole ratio of"CaCO_(3) "to" H_(2)O is 3.6 xx 10^-5`C. `"Mass of" CaCO_(3) "present in hard water" is 0.2g//L`D. `"1 miliequivalent of" CaCO_(3) "present in 1kg of hard water"` |
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Answer» Correct Answer - D |
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| 120. |
A mixture of `NH_(4)NO_(3)` and `(NH_(4))_(2)HP_(4)` coitain 30.40% mass per cent of nitrogen. What is the mass ratio of the two components in the mixture ?A. `2:1`B. `1:2`C. `3:4`D. `4:1` |
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Answer» Correct Answer - A |
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| 121. |
What weight of `NaHSO_3` is required to react with 100 " mL of " solution containing 0.33 g of `NaIO_3` according to the following reaction: `IO_3^(ɵ)+HSO_3^(ɵ)toI^(ɵ)+SO_3^(2-)` (a). `0.52g` (b). 5.2 g ltbr) (c). `1.04g` (d). `10.4g` |
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Answer» `Mw(NaHSO_3)=23+1+32+3xx16=104g` `Mw(NaIO_3)=23+127+2xx16=198g` `[6e^(-)+IO_3^(ɵ)toI^(ɵ)](n=6)` `[HSO_3^(ɵ)toI^(ɵ)](n=2)` Strength `(gL^(-1))` `=`Normality`xxEw` Normality of `NaIO_3=("Strength")/(Ew)` `=(3.3)/((198)/(6))=(3.3xx6)/(198)=0.100` Strength of `NaIO_3=(0.33g)/(100mL)` `-=(0.33xx1000)/(100)=3.3gL^(-1)` Normality of `NaIO_3=0.1N` m" Eq of "`NaIO_3=NxxV(mL)=0.1xx100=10mEq` `NaHSO_3-=NaIO_3` `mEq-=mEq-=10mEq` `m" Eq of "NaHSO_3=10mEq` `=10xx10^(-3)Eq` `=10xx10^(-3)xxEw(NaHSO_3)` `=10xx10^(-3)xx(104)/(2)g=0.52g` |
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| 122. |
Complete the following reactions: (a). `MnO_4^(2-)+H^(o+)toMn^(2+)+?` (b). `NO_2+H_2Oto?+NO` (d). `H_2O_2+Sn^(2+)toSn^(4+)+?` |
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Answer» `MnO_4^(2-)` is reduced to `Mn^(2+)` (i.e., acts as oxidising agent) and `H^(o+)` is in maximum oxidation state. So, `MnO_4^(2-)` must also be oxidised to `Mn^(7+)` state, i.e., `MnO_4^(ɵ)` will be formed. Hence, a disproportionation reaction occurs. `MnO_4^(2-)+H^(o+)toMn^(2+)toMn^(2+)+MnO_4^(ɵ)` The balanced equation is: `8H^(o+)+cancel(4e^(-))+MnO_4^(2-)toMn+4H_2O` `underline(MnO_4^(2-)toMnO_4^(ɵ)+cancel(e^(-))]xx4)` `underline(5MnO_4^(2-)+8H^(o+)toMn^(2+)+4MnO_4^(ɵ)+4H_2O)` (b). `NO_2(+4` oxidation state) disproportionates [as in (a)] to NO(+2) oxidation state) and `NO_3^(ɵ)` (`+5` oxidation state). `2H^(o+)+cancel(2e^(-))+NO_2toNO+H_2O` `underline(H_2O+NO_2toNO_3^(ɵ)+cancel(e^(-))+2H^(o+)]xx2)` (c). Here `I_2` is reduced to `2I^(ɵ)`, so `H_2O_2` must be oxidised and gives `O_2`. `cancel(2e^(-))+I_2to2I^(ɵ)` `underline(H_2O_2toO_2+2H^(o+)+cancel(2e^(-))` `underline(H_2O_2+I_2toO_2+2I^(ɵ)+2H^(o+))` (d). Here `Sn^(2+)` is oxidised to `Sn^(4+)`, so `H_2O_2` `Sn^(2+)toSn^(4+)+2e^(-)` `underline(2H^(o+)+cancel(2e^(-))+H_2O_2to2H_2O)` `underline(Sn^(2+)+H_2O_2+2H^(o+)toSn^(4+)+2H_2O)` |
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| 123. |
The `NH_(3)` evolved due to complete conversion of N from 1.12g sample of protein was absorbed in `45 "ml of" 0.4N HNO_(3)`. The excess acid required `20 "ml of" 0.1N NaOH`. The % N in the sampl is:A. 8B. 16C. 20D. 25 |
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Answer» Correct Answer - A::C |
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| 124. |
0.3 g of platinichloride of an organic diacidic base left 0.09 g of platinum on ignition. The molecular weight of the organic base isA. 120B. 240C. 180D. 60 |
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Answer» Correct Answer - B Let B is the original bese. `2B+H_2(2)PtCl_(6)toB_(2)H_(2)PtCl_(2)overset(Delta)toPt` Ew of `B_(2)H_(2)PtCl_(2)=2B+2+195+6xx35.5` `=2B+410` `("Weight of chlorophatinate")/("Weight of" Pt)=("Ew of salt")/(Ew of Pt)` `(0.3)/(0.09)=(2B+410)/(195)` `B(Ew of base)=120` Mw of `base=Ewxxacidity` `=120xx2=240` |
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| 125. |
The amount of CO in a gas ample can be determined by using the reaction. `I_(2)O_(5)+5COtoI_(2)+5CO_(2)` If a gas sample liberated 0.2 g of `I_(2)`, how many g of CO were present in the sample. |
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Answer» 1 mole of `CO=28 g.` 1 mole of `I_(2)=253.8g` 5 mole of `CO=1` mole of `I_(2)` Moles of `I_(2)=(0.2)/(253.8)` `therefore(0.2)/(253.8)` moles of `I_(2)=(5xx0.2)/(253.8)xx28g` of `CO` `=0.110 g` of `CO` |
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| 126. |
`0.50 g` of a mixture of `K_(2)CO_(3)` and `Li_(2)CO_(3)` required `30 mL` of `0.25 N HCl` solution for neutralization. What is `%` composition of mixure?A. `96%K_(2)CO_(3),4%Li_(2)CO_(3)`B. `4%K_(2)CO_(4),96%Li_(2)CO_(3)`C. `48%K_(2)CO_(3),52%Li_(2)CO_(3)`D. `52%K_(2)CO_(3),48%Li_(2)CO_(3)` |
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Answer» Correct Answer - A Mw of `K_(2)CO_(3)=39xx2+12+48=138 g mol^(-1)` `mE of K_(2)CO_(3)=(138)/(2)=69g` Mw of `LiC_(2)CO_(3)=7xx2+12+48=74 g mol^(-1)` Ew of `Li_(2)CO_(3)=(74)/(2)=37g` Let x g of `K_(2)CO_(3)` and `(0.5-x) g of Li_(2)CO_(3)` `mEq K_(2)CO_(3)+m" Eq of "Li_(2)CO_(3)=m" Eq of "HCl` `((x)/(69)+(0.5-x)/(37))xx1000=30xx0.25` `(37x+69(0.5-x))/(69xx37)=(30xx0.25)/(1000)` `69xx0.5-(30xx0.25xx69xx37)/(1000)=32x` `34.5-19.14=32x` `32x=15.36 and x=0.48` `% of K_(2)CO_(3)=(0.48xx100)/(0.5)=96%` `% of Li_(2)CO_(3)=4%` |
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| 127. |
0.7875 g of crystalline barium hydroxide is dissolved in water .For the neutralization of this solution 20 mL of N/4 `HNO_(3)` is required. How many moles of water of crystallization are present in one mole of this base ? (Given : Atomic mass Ba=137,O=16, N=14, H=1) |
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Answer» Correct Answer - 8 |
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| 128. |
If `m_(A)` gram of a metal A displaces `m_(B)` gram of another metal B from its salt solution and if the equilvalent mass are `E_(A) and E_(B)` respectively then equivalent mass of A can be expressed as:A. `E_(A)=(m_(A))/(m_(B)) xx E_(B)`B. `E_(A)=(m_(A) xx m_(B))/(E_(B)`C. `E_(A)= (m_(B))/(E_(A)) xx E_(B) `D. `E_(A)= sqrt((m_(A))/(m_(B)) xx E_(B) )` |
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Answer» Correct Answer - A |
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| 129. |
One litre of oxygen at NTP is allowed to reasonanace with three times of carbon monoxide at NTP. Calculate the volume of each gas found after the reaction. |
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Answer» The reaction equation is `underset(2"vol")(2CO)+underset(1"vol")(O_(2)) to underset(2"vol")(2CO_(2))` `"1 vol of" O_(2) "reacts with 2 vol. of CO"` or `"1 litre of" O_(2) "reacts with 2 vol. of CO"` `"Thus 1 litre of CO remains unchanged"` `"1 vol. of" O_(2) "produces" CO_(2)=2"vol"` `or "1 litre of" O_(2) "will produce" CO_(2)="2litre"` Thus, gaseous mixture after the reaction consists. Volume of CO=1litre Volume of `CO_(2)`=2 litre |
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| 130. |
What quantity of copper(II) oxide will react 2.80litre of hydrogen at NTP |
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Answer» The balanced equation is `underset("79.5g")underset(1"mol")(CuO+H_(2)) to underset(22.4"litre at NTP")underset("1mol")(Cu)+H_(2)O` 22.4 litre of hydrogen at NTP reduce CuO=79.5g 2.80 litre of hydrogen at NTP will reduce CuO `=(79.5)/(22.4)xx2.80g=9.95g` |
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| 131. |
The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by `NiCl_(2), 6H_(2)O` to form a stable coodinate compound. Assume that both the reaction are 100% complete. If 1584g of ammonium sulphate and 952g of `NiCl_(2). 6H_(2)O` are used in the preaparation the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is Atomic weight is `g mol^(-1): H-1, N=14, O=16, S=32, Cl=35.5, Ca=40, Ni=59)` |
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Answer» Reaction involved are: `(NH_(4))_(2)SO_(4)+Ca(OH)_(2) to CaSO_(4).2H_(2)O+2NH_(3)` Number of moles of `(NH_(4))_(2)SO_(4)=(1584)/(132)=12` `therefore n_(NH_(3))=24"mol" , n_(CaSO_(4).H_(2)O)=12,` `n_(NiCl_(2).6H_(2)O)=(952)/(238)=4` `underset(4"mol")(NiCl_(2).6H_(2)O)+underset(24"mol")(6NH_(3)) to underset(4"mol")([Ni(NH_(3))_(6)]Cl_(2))+6H_(2)O` `=(12xx172)+(4xx232)` =2992g |
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| 132. |
In the following reaction `As_(2)S_(3)+NO_(3)^(ɵ)+H_(2)OtoAsO_(4)^(3-)+SO_(4)^(2-)+NO+H^(o+)` The number of electrons involved in the oxidation reaction isA. 22B. 24C. 26D. 28 |
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Answer» Correct Answer - D `underset(2x=6)overset(+3)(As_(2))toundersetunderset(2x=10)(2x-16=-6)(2AsO_(4)^(3-))+4e^(-)` `underline(underset(3x=-6)(S_(3)^(2-)toundersetunderset(3x=18)(3x-24=-6)(3SO_(4)^(2-)+23e^(-)))` `underline(As_(2)S_(3)toAsO_(4)^(3-)+3SO_(4)^(2-)+28e^(-))` |
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| 133. |
n-factor of ferrous oxalate and ferric oxalate when they react with `K_(2)Cr_(2)O_(7)` is acidic medium areA. 2,6B. 6,2C. 3,6D. 6,3 |
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Answer» Correct Answer - C (i). `{:([Fe^(2+)toFe^(3+)+e^(-)),(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)):}(n=3)` (ii). `Fe_(2)(C_(2)O_(4))_(3),Fe^(3+)` ion does not change. `3C_(2)O_(4)^(2-)to6CO_(2)+6e^(-)` |
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| 134. |
if 20 mL `(M)/(10)Ba(MnO_(4))_(2)` compeletely reacts with `FeC_(2)O_(4)` in acidic medium, Q. Millimoles of `FeC_(2)O_(4)` reacted isA. `(20)/(3)`B. `(20)/(2)`C. `(20)/(6)`D. `(20)/(10)` |
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Answer» Correct Answer - A `(20)/(3)m" mol of "FeC_(2)O_(4)` `(20)/(10)=2 m" mol of "Ba(MnO_(4))_(2)` |
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| 135. |
What volume of 0.05 M `K_(2)Cr_(2)O_(7)` in acidic medium is needed for completel oxidation of 200 " mL of " 0.6 M `FeC_(2)O_(4)` solution?A. 1.2 mLB. 1.2 LC. 120 mLD. 800 mL |
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Answer» Correct Answer - B `underset(0.05M)(Cr_(2)O_(7)^(2-)+underset((0.6M)/(200mL))(FeC_(2)O_(4)overset(H^(o+))to` `Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(-)to2Cr_(3+)+7H_(2)O` `Fe^(2+)toFe^(3+)+e^(-)`, `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)` `FeC_(2)O_(4)toFe^(3+)+2CO_(2)+3e^(-)` `(0.05xx6)xxV=(0.6xx3)xx200` `impliesV=1200mL=1.2L` |
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| 136. |
if 20 mL `(M)/(10)Ba(MnO_(4))_(2)` compeletely reacts with `FeC_(2)O_(4)` in acidic medium, Q. m" Eq of "`FeC_(2)O_(4)` reacted isA. 6B. 20C. 40D. none |
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Answer» Correct Answer - B `10e^(-)+undersetunderset(2x=14)(2x-16=-2)(2MnO_(4)^(ɵ))tounderset(2x=4)(2overset(+2)(Mn))]n=10` `FeC_(2)O_(4)(n=3)` `(Fe^(2+)toFe^(2+)+e^(-), and C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))` `FeC_(2)O_(4)-=Ba^(2+)(MnO_(4)^(-1))_(2)-=CO_(2)` `C_(2)O_(4)^(2-)-=(MnO_(4)^(-2))_(2)-=CO_(2)-=20xx(1)/(10)xx10` `20mEq-=20mEq-=20mEq` |
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| 137. |
if 20 mL `(M)/(10)Ba(MnO_(4))_(2)` compeletely reacts with `FeC_(2)O_(4)` in acidic medium, Q. What is the volume of `CO_(2)` produced at `STP`A. 112 mLB. 224 mLC. 448 mLD. none |
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Answer» Correct Answer - B `1 " Eq of "CO_(2)=(22.4)/(2)L=11.2L at STP` `1000 m" Eq of "CO_(2)=11200mL at STP` `20 m" Eq of "CO_(2)=(11200)/(1000)xx20=224mL` |
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| 138. |
5 mL solution of `H_2O_2` liberates 1.27 g of iodine from an acidified KI solution. What is the molarity of `H_2O_2`? |
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Answer» `H_2O_2-=KI-=I_2` `mEq-=mEq-=mEq` `N_1xxV_1-=("weight of I_2")/(Ew(I_2))xx10^(3)mEq` `N_1xx5mL-=(1.27xx10^(3))/((127xx2)/(2))` (n-factor of `I_2=2`) `(2I^(ɵ)toI_2+2e^(-))` Normality of `H_2O_2-=2N` Molarity of `H_2O_2-=("normality")/(2)=(2)/(2)=1M` (n-factor of `H_2O_2=2)(2e^(-)+H_2O_2to2H_2O)` |
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| 139. |
10 " mL of " a gaseous hydrocarbon is exploded with 100 " mL of " oxygen. The residual gas on cooling is found to measure 95 mL, of which 20 mL is absorbed by caustic soda and the remaining by alkaline pyrogallol. The formula of the hydrocarbon is (a). `CH_4`. (b). `C_2H_6` (c). `C_2H_4` (d). `C_2H_2` |
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Answer» `C_(x)H_(y)+(x+(y)/(4))O_2toxCO_2+(y)/(2)H_2O` `1mL(x+(y)/(4))mLxmL(y)/(2)mL` `10mL10(x+(y)/(4))mL10xmL__` Volume absorbed by `KOH=` volume of `CO_2` Volume of `CO_2=20` `x=2mL=2mol` Volume absorbed by alkaline pyrogallol `=` Volume of `O_2` `=95-20=75mL` Excess of `O_2=75mL` Total `O_2=100mL` Volujme of `O_2` reacted `=100-75=25mL` `10(x+(y)/(4))=25mL` `y=2` Formula of hydrocarbon is `C_2H_2`. |
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| 140. |
When a mixture of 10 moles of `SO_(2)` and 16 moles of `O_(2)` were passed over a catalyst, 8 moles of `SO_(3)` were formed at equilibrium. The number of `SO_(2) and O_(2)` remaining unreacted were:A. 2,12B. 12,2C. 3,10D. 3,10 |
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Answer» Correct Answer - A `2SO_(2)(g)+O_(2)(g) Leftrightarrow 2SO_(3)(g)` `{:(,t=0,10,16,0),(,t_(eq),(10-2x),(16-x),0),(, ,(2x=8), ,x=4):}` `therefore "Remaining "SO_(2)=10-8=2"mol"` ` "Remaining "O_(2)=16-4=12"mol"` |
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| 141. |
Calculate the number of grams of magnesium chloride that could be obtained from 17.0g of HCl when HCl is reacted with an excess of magnesium oxide. |
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Answer» Balanced equation. `underset("1mol")(MgO)+underset(=73g)underset((2xx36.5g))underset(2"mol")(HCl) to underset(=95g)underset(2(24+71g))underset(1"mol")(MgCl_(2))+underset(1"mol")(H_(2)O)` `73"g of HCl produce" MgCl_(2)=95g` `73"g of HCl produce" MgCl_(2)=(95)/(73)g` `17"g of HCl produce" MgCl_(2)=(95)/(73)xx17g=22.12g` |
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| 142. |
An impure sample of calcium carbon contains 80% pure of impure sample reacted with excess of hydrochloric acid. Calculate the volume of carbon dioxide at NTP obtained from the sample. |
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Answer» 100g of impure of calcium carbonate contain=80g pure of calcium carbonate 25g impure calcium carbonate sample contain `=(80)/(100)xx25` =20g of pure calcium carbonate The desired equation is `underset(100g)underset(1"mol")(CaCO_(3))+2HCl to CaCl_(2)+underset("at NTP")underset(22.4"litre")(CO_(2))+H_(2)O` `100"g pure "CaCO_(3)"liberate"=(22.4)/(100)"litre"CO_(2)` `"20g pure " CaCO_(3)"liberate"=(22.4)/(100)xx20` =4.48litre `CO_(2)` |
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| 143. |
Galena (an ore) is partially oxidised by passing air through it at high temperature. After some time the passage of air stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of `O_(2)` consumed is_ , `"Atomic weights in g mol"^(-1): O =16,S=32,Pb=207` |
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Answer» Ultimate reaction involed in the extraction of lead from galena is: `PbS+O_(2) to Pb+SO_(2)` Number of moles of Pb=Number of moles of oxygen `=(1000)/(32)` Mass of lead extracted `=((1000)/(32))xx"Molar mass of lead"` `=(1000)/(32)xx207=6470g` =6.74kg |
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| 144. |
8.1 g of `K_2Cr_2O_7` reacts with 12.8 g of HI according to the equation `Cr_2O_^(2-)+HItoCrI_3+KI+I_2` Calculate: (a). Percentage by mass of `K_2Cr_2O_7` left unreasted. (b). Volume of `I_2` (g) evolved, if `I_2` obtained is heated to 500 K and 1.0 atm pressure. |
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Answer» Balance the redox reaction: `14H^(o+)+cancel(6e^(-))+Cr_2O_7^(2-)to2Cr^(3+)+7H_2O` `underline(2HItoI_2+2e^(-)+2H^(+)]xx3)` `underline(8H^(o+)+Cr_2O_7^(2-)+6HIto2Cr^(3+)+3I_2+7H_2O)` To balance equation for other ions, add `2K^(o+)` and `8I^(ɵ)` to both sides. `14HI+K_2Cr_2O_7^(2-)to2CrI_3+3I_2+7H_2O` `14xx138g of HI `requires `-=1 mol K_2Cr_2O_7-=294g` `12.8 of HI` requires`=(294xx12.8)/(14xx128)=2.1g K_2Cr_2O_7` (a). Weight of `K_2CrO_2O_7` unreacted `=8.1-2.1=6.0g` `%` of `K_2Cr_2O_7` unreacted`=(6xx100)/(8.1)=74.07%` (b). 1 " mol of "`K_2Cr_2O_7(=294g) gives =3 " mol of "I_2` `2.1 g of K_2Cr_2O_7-=(3xx2.1)/(294)=0.021 mol I_2` `PV=nRT` `V_(I_2)=(nRT)/(P)=(0.021xx0.082xx500)/(1)` `=0.861L=861 mL` |
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| 145. |
N-factor for the following reaction is `FeS_(2)toFe_(2)O_(3)+SO_(2)`A. 8B. 9C. 10D. 11 |
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Answer» Correct Answer - D `overset(+2)(Fe)overset(-1xx2)/(S_(2))tooverset(+3xx2)(Fe_(2))overset(-2xx3)(O_(3))+overset(+4-2xx2)(SO_(2))` `underset(2x=4)(2Fe^(2+))overset(0)tounderset(2x=6)(Fe_(2)^(3+)+2e^(-)]n=(2)/(2)=1` `{:(underset(2x=-2)(S_(2)^(2-))toundersetunderset(2x=18)(2x-8=0)(2SO_(2))+10e^(-)]n=10` Total n factor `=10+1=11` |
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| 146. |
Which of the following statements is/are correct about the followig reaction? `Fe_(3)O_(4)overset(Delta)toFe_(2)O_(3)`.A. The equivalent weight of `Fe_(3)O_(4)` is `M_(1)(M_(1)=` molecular weight of `Fe_(2)O_(4))`B. The equivalent weight of `Fe_(3)O_(4)` is `(M_(1))/(3)`.C. The equivalent weight of `Fe_(2)O_(3)` is `(3M_(2))/(2)(M_(2)=` molecular weight of `Fe_(2)O_(3))`.D. The equivalent weight of `Fe_(2)O_(3)` is `(M_(2))/(2)`. |
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Answer» Correct Answer - A::C `{:(Fe_(3)O_(4),Fe_(2)O_(3)),(3x-8=0(3x=8),2x-6=0 (2x=6)),(x=(8)/(3),x=3):}` `_(2)(Fe^((8)/(3)+))_(3)to3(Fe^(3+))_(2)+2e^(-)` `{:[(x=(8)/(3)xx2xx3),(x=16),("n-factor"=(2)/(2)=1),(Ew_(1)=(M_(1))/(1))]:}` `{:[(x=18),(3mol=2e^(-)),(1mol=(2)/(3)e^(-)),("n-factor"=(3)/(2)),(Ew_(2)=(M_(2))/((2)/(3))=(3M_(2))/(2)):}]` |
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| 147. |
When `KIO_(3)` solution is heated with excess of oxalic acid it is found that 1.683 g of `KIO_(3)` is consumed per gram of iodine liberated. Formulate the stoichemistry of the products. (Atmic weight of iodine`=127 and K=-39`) |
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Answer» Knowing the products, we formulate the stoichiometry and then check its correctness, `2KIO_(3)+6H_(2)C_(2)O_(4)toK_(2)C_(2)O_(4)+underset(254g)(I_(2))+6H_(2)O+10CO_(2)` As per this equation 1 g of iodine corresponds to `(428)/(254)=1.685g`, which agrees with the given data. |
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| 148. |
`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. 1 molB. 0.5 molC. 0.25 molD. 2 mol |
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Answer» Correct Answer - C `I^(ɵ)` is limiting reagent so one mole `I^(ɵ)` will give `(1)/(4)` mol or `0.25` moles of `Hgl_(4)^(2-)` |
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| 149. |
Calculate the mass of oxalic acid `(H_(2)C_(2)O_(4))` which can be oxidised to `CO_(2)` by `100.0mL` of `MnO_(4)^(-)` solution, `10 mL` of which is capable of oxidising `50.0mL` of `1.0NI^(-)` to `I_(2)`?A. 45gB. 22.5gC. 30gD. 12.25g |
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Answer» Correct Answer - B |
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| 150. |
100 " mL of " `(M)/(10)` `Ca(MnO_(4))_(2)` in acidic medium can be oxidised completely withA. `100 " mL of " 1 M FeSO_(4)` solutionB. `(100)/(3) " mL of " 1 M FeC_(2)O_(4)` solutionC. 25 " mL of " 1 M `K_(2)Cr_(2)O_(7)` solutionD. 75 " mL of " 1 M `C_(2)O_(4)^(2-)` solution |
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Answer» Correct Answer - A::B `(10e^(-)+2MnO_(4)^(ɵ)to2Mn^(2+))(n=10)` (a). `Fe^(2+)toFe^(3+)+e^(-)(n=1)` m" Eq of "`Ca(MnO_(4))_(2)=100xx(1)/(10)xx10=100` m" Eq of "`FeSO_(4)=100xx1=100` (b). `m" Eq of "FeC_(2)O_(4)(n=3),{:[(Fe^(2+)toFe^(+3)+e^(-)),(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))]:}` `=(100)/(3)xx3xx1` `=100=m" Eq of "Ba(MnO_(4))_(2)` (c). `m" Eq of "K_(2)Cr_(2)O_(7)(n=6)=25xx1xx6=150` (d). `m" Eq of "C_(2)O_(4)^(2-)(n=2)=75xx2xx1=150`. |
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