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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.
| 201. |
Two elements `A` and `B` combine chemically to from compounds combining with a fixed mass of `A` in I, II and III is `1:3:5`, if 32 parts by mass of `A` combine with 84 parts by mass of `B` in II, then III, 16 parts of `A` will combine with................ by mass of `B`.A. 14 parts by mass of YB. 42 parts by mass of YC. 70 parts by mass of YD. 84 parts by mass of Y |
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Answer» Correct Answer - C |
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| 202. |
STATEMENT-1 : `H_(2)SO_(4)` can not act as reducing agent. STATEMENT-2 : Sulphur can not increase its oxidation number beyond `+6`.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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| 203. |
Assertion: Fluorine exists only in `-1` oxidation state. Reason: Fluorine has `2s^(2)2p^(5)`configuration.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 - B |
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| 204. |
STATEMENT-1 : `I_(2)rarrIO_(3)^(-)+I^(-)`, is example of a disproportionation reaction. STATEMENT-2 : Oxidation number of I can vary from `-1` to `+7`.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 - B |
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| 205. |
STATEMENT-1: In `CrO_(5)` oxidation number of Cr is +6. STATEMENT-2 : `CrO_(5)` has butterfly structure in which peroxide peroxide bonds are present. 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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| 206. |
STATEMENT-1 : `[Fe(CN)_(6)]^(4) to Fe^(3+)+CO_(2)+NO_(3)^(-)`, the equivalent mass of reactant is 3.74. STATEMENT-2 : "Equivalent mass of reactant" = ("Mol.mass")/(61)`.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 - D |
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| 207. |
STATEMENT-1 : In the given reaction, `NaOH+H_(3)PO_(4)rarr NaH_(2)PO_(4)+H_(2)O` equivalent mass of `H_(3)PO_(4)` is `M//3` STATEMENT-2 : `H_(3)PO_(4)` is tribasic acid.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 - D |
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| 208. |
When potassium permanganate is titrated against ferrous ammonoum sulphate, the equivalent weight of potassium permanganent isA. `("molecular mass")/(3)`B. `("molecular mass")/(5)`C. `("molecular mass")/(2)`D. `("molecular mass")/(10)` |
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Answer» Correct Answer - B |
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| 209. |
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` many moles `NaClO_(3)` obtained after the complection of reaction by taking 1 mole `Cl_(2)` and other reagents in excess ?A. `(1)/(3)` moleB. ZeroC. `(1)/(4)` moleD. 1 mole |
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Answer» Correct Answer - B |
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| 210. |
What mass of `H_(2)C_(2)O_(4). 2H_(2)O (mol.mass=126)` should be dissoved in water to prepare 250mL of centinormal solution which act as a reducing agent?A. 0.63gB. 0.1575gC. 0.126gD. 0.875g |
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Answer» Correct Answer - B |
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| 211. |
A `2.0 g` of mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` loses `0.248 g` when heated to `300^(@)C`, the temperature at which `NaHCO_(3)` decomposes to `Na_(2)CO_(3)`, `CO_(2)` and `H_(2) O`. What is the percentage of `Na_(2) CO_(3)` in mixture? |
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Answer» `underset(168g)(2Na HCo_(3))to Na_(2) CO_(3)+ underset(62g)underset( ubrace(44" "18))(CO_(2)+H_(2)O` The loss comes due to evolution of `CO_(2)` and steam 6.2 g loss occurs when the quantity of `NAHCO_(3)` is 168 g. 0.248g loss will ocuur when the quantity of `NaHCO_(3) = (168)/(62) xx 0.248 = 0.672 g` Quanity of `Na_(2) CO_(3)` in the sample `= (2.0 - 0.672) = 1.328 g` % of `Na_(2)CO_(3) = (1.328)/(2) xx 100 = 66.4` |
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| 212. |
A mixture of `NaHCO_(3) and Na_(2)CO_(3)`, weighed 1.0235. The dissolved mixture was reached with excess of `Ba(OH)_(2)` to form 2.1028g `BaCO_(3)`, by the following reactions: `Na_(2)CO_(3)+Ba(OH)_(2)toBaCO_(3)+2NaOH` `NaHCO_(3)+Ba(OH)_(2) to BaCO_(3)+NaOH+H_(2)O` What was the percentage of `NaHCO_(3)` in the orginal mixture? |
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Answer» Let x g of `NaHCO_(3)` be present in the mixture? Mass of `Na_(2)CO_(3)` in the mixture=(1.025-x)g` Number of moles of `NaHCO_(3)=(x)/(84)` Number of moles of `Na_(2)CO_(3)`=(1.0235-x)/(106)` Number of moles of `BaCO_(3)`=Number of moles of `NaHCO_(3)+"Number of moles of" Na_(2)CO_(3)` `(2.1028)/(197)=(x)/(84)+(1.0235-x)/(106)` x=0.4122 `"Amount of " NaHCO_(3)=0.4122g` `"Percentage of "NaHCO_(3)=(0.4122)/(1.0235)xx100=40.27` |
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| 213. |
100 ml of gaseous mixture containing `CO, CO_(2) and O_(2)` was sparked there was contraction of 80mL volume when the nixturc was passed through aqueous caustic potash KOH. The omposition of initial gaseous mixture will be respectively: |
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Answer» Correct Answer - c,d `CO+(1)/(2)O_(2) to CO_(2)` 10mL `O_(2)` will give 20mL `CO_(2)` in above reaction. Thus, total volume of `CO_(2)` will be 80mL which will be absorbed in caustic potash KOH. Similarly in (c ) O option, 30mL `CO_(2)` will be formed on sparking. `underset("Limiting")underset("30mL")(CO)+(1)/(2)O_(2) to underset("30mL")(CO_(2))` Total volume of `CO_(2)` will be 80mL |
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| 214. |
Calculate the amount of lime that can be produced by heating 100g to 90% pure limestone |
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Answer» Correct Answer - a,c `underset(100g)(CaCO_(3)) to underset(56g)(CaO)+CO_(2)` `therefore "100g of "CaCO_(3)-=56"g "CaO` `therefore "90g "CaCO_(3)-=(56)/(100)xx90"g "CaO=(50.4)/(56)=0.9` |
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| 215. |
For the reaction `M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)` if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction isA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A `M^(x+)+MnO_(4)^(ɵ)toMO_(3)^(ɵ)+Mn^(2+)` `MnO_(4)^(ɵ)+8H^(o+)+5e^(-)toMn^(2+)+4H_(2)O` `M^(x+)+3H_(2)OtoMO_(3)^(ɵ)+6H^(o+)+(5-x)e^(-)` `m" Eq of "MnO_(4)^(2-)-=m" Eq of "M^(x+)` `1xx5-=[1.67xx(5-x)]implies5-x=3impliesx=2` |
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| 216. |
`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution. |
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Answer» For `H_(2)SO_(4)`, n factor`=2` Let there be `n m" Eq of "H_(2)SO_(4)`. m" Eq of "all acids in `2L=(5xx8)+(48.5xx5)+n` `=64+n` Normality`=((6n+n))/(2000)` Normality of `Na_(2)CO_(3).10H_(2)O` is `(1)/((286)/(2))xx(1)/(100)xx1000=0.07` `thereforeN_(1)V_(1)=N_(2)V_(2)` or `((64+n))/(200)xx30=0.07xx42.9` or `n=136.2 m" Eq of "H_(2)SO_(4)` Moles of `H_(2)SO_(4)=(136.2)/(2)xx10^(-3)` Mass of `SO_(4)^(2-)=(136.2)/(2)xx10^(-3)xx96` `=6.5376g` |
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| 217. |
What volume of 0.1 M `Ba(OH)_(2)` will be required to neutralise a mixture of 50 " mL of " 0.1 M HCl and 100 " mL of " 0.2 M `H_(3)PO_(4)` using methyl red indicator?A. 25 mLB. 50 mLC. 100 mLD. 125 mL |
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Answer» Correct Answer - D Methyl red indicates complete ionisation of `HCl` `(n=1)` and first step ionisation of `H_(3)PO_(4)` (`n=1`) First case: `Ba(OH)_(2)-=HCl` `N_(1)V_(1)=N_(2)V_(2)` `0.1xx2xxV_(1)=0.1xx1xx50` `V_(1)=25mL` Second case: `Ba(OH)_(2)-=H_(2)PO_(4)(n=1)` `0.1xx2xxV_(1)=0.2xx1xx100` Total volume `=100+25=125mL` |
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| 218. |
What volume of 0.2 M KOH will be requried to neutralise 100 " mL of " 0.1 M `H_(3)PO_(4)` using methyl red indicator (change of colour pink `rarr` yellow) and then bromothymol blue indicator is added.A. 50 mLB. 100 mLC. 150 mLD. 200 mL |
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Answer» Correct Answer - B Methyl red indicates the first step ionisation of `H_(2)PO_(4)` bromothymol blue indiates the second step ionisation of `H_(3)PO_(4)`, i.e., `H_(2)PO_(4)^(ɵ)toH^(o+)+HPO_(4)^(2-)` `(n=1)(N=Mxx1)` First case: When methyl red is added (change of `H_(2)PO_(4)toH_(2)PO_(4)^(ɵ)+H^(o+))(n=1)(N=Mxx1)` `KOH-=H_(3)PO_(4)` `N_(1)V_(1)-=N_(2)V_(2)` `0.2xx1xxV_(1)=0.1xx1xx100` `V_(1)=50mL` Second case: When bromothymol blue is added (change of `H_(2)PO_(4)^(ɵ)toH^(o+)+HPO_(4)^(2-))` `(n=1)(N=Mxx1)` `KOH-=H_(2)PO_(4)^(ɵ)` `N_(1)V_(1)-=N_(2)V_(2)` `0.2xx1xxV_(1)=0.1xx1xx100` `V_(1)=50mL` Total volume `=50+50=100mL` |
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| 219. |
Ratio of moles of Fe (II) oxidised by equal volumes of equimolar `KMnO_(4)` and `K_(2)Cr_(2)O_(7)` solutions in acidic medium will be:A. `5:3`B. `1:1`C. `1:2`D. `5:6` |
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Answer» Correct Answer - D |
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| 220. |
`0.45 g` of acid (mol. Wt.`=90`) was exactly neutralized by `20 ml` of `0.5(M) NaOH`. The basicity of the given acid isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B |
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| 221. |
There are three acid-base indicators. Methyl orange (end point at `pH=4`), bromothymol blue (end point at `pH-7`), phenolphthalein (end point at `pH=9`). Which is the most suitable indicator for the following titrations? (a). `H_2SO_4` with KOH (b). KCn with HCl (c). `NH_3` with `HNO_3` (d). HF with NaOH |
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Answer» (a). Strong acid and strong base: any indicator can be used. (b). Strong acid and salt of weak acid and strong base: Methyl orange is suitable. (c). Weak base and strong acid: Methyl orange (d). Strong base and weak acid: Phenolphthalein. |
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| 222. |
A mixtures of methane and ethylene in the volume ration x:y has total volume of 30mL. On complete combustion it gave 40mL of `CO_(2)`. IF the ratio of original mixture is y:x instead of x:y. What volume of `CO_(2)` would have been produced a combustion?A. 50mLB. 75mLC. 100mLD. 125mL |
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Answer» Correct Answer - A Combustion of `CH_(4) and C_(2)H_(4)` may be represented as. `"Combustion of "CH_(4) "may be represented as",` `underset(xmL)(CH_(4)(g))+2O_(2)(g) to underset(xmL)(CO_(2)(g))+2H_(2)O(g)` `underset(ymL)(C_(2)H_(4)(g))+3O_(2)(g) to underset(ymL)(2CO_(2)(g))+2H_(2)O(g)` Given, x+y=30 x+2y=40 `therefore x=20, y=10` When we take y mL `CH_(4) and "x mL" C_(2)H_(4)`, then volume of `CO_(2)` will be (y+2x), i.e. 50mL. |
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| 223. |
A water is said to be soft water if it produces sufficient foam with the soap and water that does not produce foam with soap is known as hard water. Hardness has been classified into two types (i)Temporary hardness (ii) Permanent hardness. Temporary hardness is due to presence of calcium and magnesium bicarbonate. It is simply removed by boiling as `Ca(HCO_(3))_(2)overset(Delta)rarr CaCO_(3)darr+CO_(2)uarr+H_(2)O` `Mg(HCO_(3))_(2)overset(Delta)rarr MgCO_(3)darr+CO_(2)uarr+H_(2)O` temporary hardness can also be removed by addition of slaked lime, `Ca(OH)_(2)` `Ca(HCO_(3))_(2)+Ca(OH)_(2)to2CaCO_(3)darr+2H_(2)O` permanent hardsness is due to presencce of sulphates and chlorides of Ca,Mg,etc. It is removed by washing soda as `CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)darr+2NaCl` `CaSO(4)+Na_(2)CO_(3)to CaCO_(3)darr+Na_(2)SO_(4)` Permanent hardness also removed by ion exchange resin process as `2RH+Ca^(2+) to R_(2)Ca+2H^(+)` `2ROH+SO_(4)^(2-)toR_(2)SO_(4)+2OH^(-)` The degree of hardness of water is measured in terms of PPm of `CaCO_(3)` 100 PPm means 100 g of `CaCO_(3)` is present in `10^(6)` g of `H_(2)O`. If any other water sample which contain 120 PPm of `MgSO_(4)`, hardness in terms of `CaCO_(3)` is equal to =100 PPm. What is the mass of `Ca(OH)_(2)` required for 10 litre of water remove temporary hardness of 100 PPm due to `Ca(HCO_(3))_(2)` ?A. 1.62 gB. 0.74 gC. 7.4 gD. None of these |
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Answer» Correct Answer - B |
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| 224. |
30g Mg and 30g `O_(2)` are reacted and the residual mixture contains:A. 60g of MgO onlyB. 40g of MgO and 20g of `O_(2)`C. 45g of MgO and 15g of `O_(2)`D. 50g of MgO and 10g of `O_(2)` |
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Answer» Correct Answer - D `underset(2xx24g)underset(2"mol")(2Mg(s))+underset(1xx32g)underset(1"mol")(O_(2)(g)) to underset(2xx40g)underset(2"mol")(2MgO(g))` `30"g Mg gives "((80)/(48)xx30)"` g MgO on complete reaction, i.e. 30gMg=50g MgO `30g O_(2)" gives "((80)/(48)xx30)"` MgO on complete reaction, i.e.=30g `O_(2)-=75"g MgO"` Mg is limiting reactant and MgO formed in the reaction will be 50g. Unreacted amount of `O_(2)`=30-20=10g Mixture contians 50g of MgO and 10g `O_(2)`. |
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| 225. |
In the determination of hardness of a sample of water, the following results were obtained: Volume of sample of `H_2O=100mL` Volume of `(N)/(50)Na_2CO_3` added to it `=20mL` Volume of `(N)/(50)H_2SO_4` used to back titrate the unreacted `Na_2CO_3=10mL` Calculate the hardness of water in g `L^(-1)` |
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Answer» m" Eq of "unreacted `Na_(2)CO_(3)=m" Eq of "H_(2)SO_(4)` `=(1xx10)/(50)=(1)/(5)` Total m" Eq of "`Na_(2)CO_(3)` taken `=(1xx20)/(50)=(2)/(5)` m" Eq of "`Na_(2)CO_(3)` absorbed by `H_(2)O=(2)/(5)-(1)/(5)=(1)/(5)` Weight of `Na_(2)CO_(3)` absorbed by `H_(2)O` `=(1)/(5)xx10^(-3)xx53g (Ew_(Na_(2)CO_(3))=53` `=0.0106 g ((Na_(2)CO_(3))/(100mL)) of H_(2)O` `=(100xx0.0106)/(106)g of CaCO_(3)` `=0.01 g of CaCO_(3)` `=(0.1xx1000)/(100)"g of "((CaCO_(3))/(LH_(2)O))` `=0.1g" of "(CaCO_(3))/(L)" of "H_(2)O` |
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| 226. |
(a). What is the equivalent weight of (i) `Fe(HC_2O_4)_2`, (b). Equivalent weight of `Cl_2` is 42.6 in the following disproportionation reaction: `Cl_2+overset(ɵ)(O)HtoCl^(ɵ)+H_2O+?` (oxidised product) identify the oxidised product. (c). What is the equivalent weight of `K_2S_2O_3` in the reaction with `I_2` in acidic medium? |
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Answer» (a). (i). Ew of `Fe(HC_2O_4)_2^(-1xx2)` as reducing agent `Fe^(2+)` is oxidised to `Fe^(3+)` and `2C_2O_4^(2-)` to `4CO_2` `{:(Fe^(2+)toFe^(3+)+e^(-)),(2C_2O_4^(2-)to4CO_2+4e^(-)):}]n=5` Mw of `Fe(HC_2O_4)_2` as an acid. There are `2H^(o+)` `Ew=(234)/(2)=117g` (ii). Ew of `Fe(HC_2O_4)_3^(-1xx3)` as reducing agent. `Fe^(3+)` does not oxidise only `3C_2O_4^(2-)` is oxidised to `6CO_2` `3C_2O_4^(2-)to^CO_2+6e^(-)` Mw of `Fe(HC_2O_4)_3` `=56+3(1+2xx12+4xx16)=323g` `Ew=(234)/(3)=78.0g` (b). `2e^(-)+Cl_2to2Cl^(ɵ)` (reduction) `Cl_2` can be oxidised eithr to `ClO^(ɵ)` (in cold) or `ClO_3^(ɵ)` (in hot). Since condition is not given, so check which is formed. `{:(Cl_2to2ClO^(ɵ)+2e^(-)):}` `Ew=((Mw)/(2)+(Mw)/(2))=(71)/(2)+(71)/(2)=71g` `{:(Cl_2to2ClO_3^(ɵ)+10e^(-)):}` `Ew=((Mw)/(2)+(Mw)/(10))=(71)/(2)+(71)/(10)` `=35.5+7.1=42.6` Given `Ew=42.6, so ClO_3^(ɵ)` is formed. (c). Mw of `K_2S_2O_3=2xx39+2xx32+3xx16=190` In acidic medium `2S_2O_3^(2-)toS_4O_6^(2-)+2e^(-)(n=(2)/(2)=1)` `Ew=(190)/(1)=190.0g` |
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| 227. |
`P_4` undergoes disproportionation in basic medium to give `PH_3` (phosphine) and `H_2PO^(ɵ)` (dihydrogen hypophoshite ion). Atomic weight of P is 31. |
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Answer» `P_4to4H_2PO_2^(ɵ)+4e^(-)(n=4)` (oxidation) `12e^(-)+P_4to4PH_3(n=12)` (reduction) `Ew(P_4)=(Mw(P_4))/(4)+(Mw(P_4))/(12)` `=(31xx4)/(4)+(31xx4)/(12)` `=31+(31)/(3)=31+10.33=41.33g` Alternatively, `Ew(P_4)=Mw(P_4)((1)/(4)+(1)/(12))=Mw(P_4)xx(1)/(3)` `:.` n-factor of `P_4=3` second method : `P_(4)to4H_(2)PO_(2)^(ɵ)+cancel(4e^(-))]xx3` `underline(12^(-)+P_(4)to4PH_(3))` `underline(4P_(4)to4H_(2)PO_(2)^(ɵ)+4PH_(3))` Total change in number of electrons `=12` Effective molecular weight of `P_(4)=Mw(P_(4))+3Mw(P_(4))` `=Mw(P_(4))` `Ew(P_(4))=(4Mw(P_(4)))/(12)=(4xx4xx31)/(12)=41.33`g `therfore` n-factor of `P_(4)=(12)/(4)=3` |
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| 228. |
Metalic tin (Sn) is oxidised to its maximum oxidation state by `KMnO_4` and `K_2Cr_2O_7` separately in the presence of HCl. Calculate the ratios of the volumes of decimolar solutions of `KMnO_4` and `K_2Cr_2O_7` that would be reduced by 1.0 g of Sn (Atomic weight of `Sn=118.6`). |
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Answer» Sn is oxidised to `+4` oxidation state (maximum oxidation state of `Sn=+4`) `SntoSn^(+4)+4e^(-)(n=4)` `[5e^(-)+MnO_4^(ɵ)toMn^(2+)](n=6)` Equivalent od `Sn=("Weight")/(Ew)=(1.0)/((118.6)/(4))=(4)/(118.6)Eq` `Sn-=MnO_4^(ɵ)` `Eq-=Eq` `(4)/(118.6)-=0.1xx5` (n-factor)`xxV(L)` `V_(MnO_4^(ɵ))=(4)/(118xx0.1xx5)=(4)/(59.3)=0.067L` `Sn-=Cr_2O_7^(2-)` `Eq-=Eq` `(4)/(118.6)-=0.1xx6` (n-factor)`xxV(L)` `V_(Cr_2O_7^(2-))=(4)/(118.6xx0.1xx6)=(4)/(71.16)=0.057L` Alternatively `(V_(MnO_4^(ɵ)))/(V_(Cr_2^(2-)))=(6)/(5)=6:5(("f-factor " Cr_2O_7^(2-))/("n-factor of "MnO_4^(ɵ)))` Since the same equivalent of Sn is reating with 0.1 M of `KMnO_4` and `K_2Cr_2O_7`, the ratios of the volumes of `KMnO_4` and `K_2Cr_2O_7` reacting with the same equivalent of Sn is inversely proportional to their n-factors. |
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| 229. |
In a chrome plating plant, `CrO_4^(2-)` (chromate) ions are present in waste water. The chromate ions are reduced to insoluble chromium hydroxide, `Cr(OH)_3`, by dithionate ion, `S_2O_4^(2-)` in basic medium. `CrO_4^(2-)+S_2O_4^(2-)+overset(ɵ)(O)H+H_2OtoCr_2(OH_3)+SO_3^(2-)` 10 L of water requires 522 g of `Na_2S+2O_4`. Calculate the normality and molarity of `CrO_4^(2-)` in waste water. Also express the concentration of `Na_2CrO_4` in ppm. |
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Answer» `{:[undersetunderset(2x=6)(x-8=-2)(S_2O_4^(2-))toundersetunderset(2x=8)(2x-12=-4)(2SO_3^(2-))+2e^(-):}](n=2)` `{:[undersetunderset(x=6)(x-8=-2)(CrO_4^(2-)+3e^(-))toundersetunderset(x=3)(x-3=0)(Cr(OH))]:}(n=3)` `Mw(Na_2S_2O_4)=2xx23+2xx32+16xx4=174g` `(Ew=(174)/(2))` `S_2O_4^(2-)-=CrO_4^(2-)` `Eq-=Eq` `("Weight")/("Equivalent weight")-=Eq` `(522)/((174)/(2))-=Eq` `therefore` 6 " Eq of "`S_2O_4^(2-)-=(6" Eq of "CrO_4^(2-))/(10L of "water")` `-=(6)/(10)=0.6EqL^(-1)` `therefore` Normality of `CrO_4^(2-)=0.6N` Molarity of `CrO_4^(2-)=(N)/("n-factor")=(0.6)/(3)=0.2M` Strength of `Na_2CrO_4=MxxMw` `(Mw Na_2CrO_4=2xx23+52+16xx4=162g)` `0.2xx162` `=32.4gL^(-1)` Concentration in ppm `=("Weight of" Na_2CrO_4" in "1000 mL xx10^(6))/(1000mL)` `=(32.4xx10^(6))/(10^(3))=32400p p m` |
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| 230. |
Equivalent weight of `H_(3)PO_(2)` when it disproportionates into `PH_(3)` and `H_(3)PO_(3)` is (mol.wt. of `H_(3)PO_(2)=M`)A. MB. `(M)/(2)`C. `(M)/(4)`D. `(3M)/(4)` |
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Answer» Correct Answer - D |
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| 231. |
Equivalent weight of `H_(3)PO_(2)` when it disproportionates into `PH_(3)` and `H_(3)PO_(3)` is (mol.wt. of `H_(3)PO_(2)=M`)A. `M`B. `(M)/(2)`C. `(M)/(4)`D. `(3M)/(4)` |
| Answer» Correct Answer - D | |
| 232. |
`SO_(2)` gas is slowly passed through an aquesous suspension containing 12g `CaSO_(3)` till the milkiness just disappears, what amount of `SO_(2)` would be required? |
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Answer» Correct Answer - d,d The reaction involed is `underset("64g")underset("1mol")(underset("120g")underset("1mol")underset("milky")(CaSO_(3))+H_(2)O+SO_(2) to underset("colourless")(Ca(HSO_(3))_(2)))` `12"g "CaSO_(3) "will react with" 6.4g"SO_(2) or 0.1"mol" SO_(2)` |
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| 233. |
A 4.0 g sample contained `Fe_2O_3,Fe_3O_4`, and inert material. It was treated with an excess of aq KI solution in acidic medium, which reduced all iron to `Fe^(2+)` ions. The resulting solution was diluted to 50 mL and a 10 mL sample of it was taken the iodine liberated in the small sample was titrated with 12.0 " mL of " 0.5 M `Na_2S_2O_3` solution. The iodine from another 25 mL was extracted, after which the `Fe^(2+)` ions were titrated with 16 " mL of " 0.25 M `MnO_4^(ɵ)` ions in `H_2SO_4` solution. Calculate the mass of two oxides in the original mixture. |
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Answer» Let `Fe_2O_3-=xmmol` `Fe_3O_4-=ymmol-=FeO+Fe_2O_3` `Fe^(3+)=2(x+y)mmol and Fe^(2+)=y m mol` `2Fe^(3+)+2I^(ɵ)toI_2+2Fe^(2+)` `2m" mol of "Fe^(3+)-=1 mmol I_2` `mmol I_2-=2m" mol of "S_2O_3^(2-)` (in 20 mL sample) `(x+y)(10)/(50)=(1)/(2)xx12xx0.5` `impliesx+y=15` Note that now m" mol of "`Fe^(2+)` in 50 mL sample is `2(x+y)+y=2x+3y` Also, `5Fe^(2+)+MnO_4^(ɵ)to5Fe^(3+)Mn^(2+)` `5m" mol of "Fe^(2+)-=1 m" mol of "MnO_4^(ɵ)` `implies(25)/(50)(2x+3y)=5(16xx0.25)` `implies2x+3y=40` Solving, we get `x=5 and y=10` Mass of `Fe_2O_3=(5)/(1000)xx160=0.8g` Mass of `Fe_2O_4=(10)/(1000)xx232=2.32g` |
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| 234. |
1.0 g of metal nitrate gave 0.86 g of metal carbonate. Calculate the Equivalent weight of metal. |
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Answer» Let the equivalent weight of metal `=E` Ew of metal nitrate`=Ew of M+EwNO_3^(ɵ)=E+62` Equivalent weight of metal carbonate `=` Equivalent weight of metal `+` Equivalent weight of `CO_3^(2-)` `=E+(60)/(2)=E+30` `therefore("Ew of M nitrate")/("Ew of M carbonate")=("Weight of M nitrate")/("weight of M sulphate")` `(E+62)/(E+30)=(1.0)/(0.86)` `thereforeE=166.57` |
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| 235. |
1 mole of methylamine on reaction with nitrous acid gives at NTP:A. 1 litre of nitrogenB. 22.4 litre of nitrogenC. 11.2 of nitrogenD. 5.6 litre of nitrogen |
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Answer» Correct Answer - B `underset("1 mol")(CH_(3))- NH_(2)+ HONO to CH_(3) - OH +underset("22.4 litre at STP")(N_(2)) + H_(2)O` |
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| 236. |
The oxidation number of nitrogen atoms in `NH_(4)NO_(3)` are:A. `+3,+3`B. `+3,-3`C. `-3,-5`D. `-5,+3` |
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Answer» Correct Answer - C |
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| 237. |
250 " mL of " x M solution and 500 " mL of " y M solution of a solute are mixed and diluted to 2L to produce a final concentration of 1.6 M. If `x:y=5:4`, calculate x and y. |
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Answer» (iii). `250 mLxx x+500 mL xxy=1.6Mxx2000mL` `x+2y=12.8` ……(i) `(x)/(y)=(5)/(4)` `4x=5y` `x=(5)/(4)y` putting the value of x I equation (i), we get `(5)/(4)y+2y=12.8` `x=4.93M,y=3.94M` |
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| 238. |
`N_(2)`(g) reacts with `H_(2)`(g) in either of the following ways depending upon supply of `H_(2)`(g) : `N_(2)(g)+H_(2)(g) to N_(2)H_(2)(l)` `N_(2)(g)+2H_(2)(g) to N_(2)H_(4)`(g) If 5 L `N_(2)`(g) and 3 L `H_(2)`(g) are taken initially (at same temperature and pressure ), calculate the contraction in valume after the reaction (in L ). |
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Answer» Correct Answer - 6 |
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| 239. |
200 mL of 1 HCl, is mixed with 300 mL of 6 M and the final solution is diluted to 1000 mL.calculate molar concentration of `[H^(+)`] ion . |
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Answer» Correct Answer - 2 |
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| 240. |
`100 mL` of ozone at `STP` was passed through `100 mL` of `10` volume `H_(2)O_(2)` solution. What is the volume strength of `H_(2)O_(2)` after attraction?A. 9.5B. 9C. 4.75D. 4.5 |
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Answer» Correct Answer - A `O_(3)toO_(2)+O` .(i) `H_(2)O_(2)toH_(2)O+O` ..(ii) `underset((1)/(2)vol)(O)+underset((1)/(2)vol)(O)tounderset(1 vol)(O_(2))` ..(iii) From equations (i) and (ii), we inter that 100 " mL of " `O_(3)` at STP will produce 100 " mL of " molecular `O_(2)` as such and 100 " mL of " oxygen molecular after reaction with `H_(2)O_(2)`. This new volume of 100 " mL of " molecular oxygen after reaction with `H_(2)O_(2)` is contributed equally by `O_(3)` and `H_(2)O_(2)`. Thus, 50 " mL of " oxygen have been contributed by `H_(2)O_(2)`. Again we know Volume of `H_(2)O_(2)xx` volume stregnth of `H_(2)O_(2)` `=` Volume of `O_(2)` at STP `therefore100 " mL of " 10 V H_(2)O_(2)-=1000" mL of " O_(2)` STP After utilisation of 50 " mL of " `O_(2)`, according to equation (iii), the balance `(1000-50)=950mL` of `O_(2)` at STP are still retainable by `100" mL of " H_(2)O_(2)`. Hence volume strength of `H_(2)O_(2)` after reaction `=("Volume" of O_(2) at STP)/("volume" of H_(2)O_(2))=(950)/(100)=9.5V` `therefore` Volume strength `=9.5` |
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| 241. |
Fe show on oxidation state of +1 in:A. `Fe(CO)_(5)`B. `[Fe(H_(2)O)_(5)NO]SO_(4)`C. `Fe_(4)[Fe(CN)_(6)]_(3)`D. `Fe_(4)Cl_(4)^(-)` |
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Answer» Correct Answer - B |
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| 242. |
`3.4 g` sample of `H_(2)O_(2)` solution containing `x% H_(2)O_(2)` by weight requires `x mL of a KMnO_(4)` solution for complete oxidation under acidic condition. The normality of `KMnO_(4)` solution isA. 1NB. 2NC. 3ND. 0.5N |
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Answer» Correct Answer - B 100 g of `H_(2)O_(2)` sample solution contains x g of `H_(2)O_(2)`. 3.4 g of solution contains`=(x)/(100)xx3.4` Weight of `H_(2)O_(2)=(3.4x)/(100)` " Eq of "`H_(2)O_(2)=(3.4x)/(100)xx(1)/(17)` `m" Eq of "H_(2)O_(2)=(3.4x)/(100xx17)xx1000=(34x)/(17)=2x` `m" Eq of "KMnO_(4)=x xxN` `x xxN=2x` `N=2` |
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| 243. |
10 " mL of " a solution of `H_(2)O_(2)` of 10 violume strength decolourises 100 " mL of " `KMnO_(4)` solution acidified with dil `H_(2)SO_(4)`. The amount of `KMnO_(4)` in the given solution is `K=39, Mn=55)`A. 0.282 gB. 0.564 gC. 1.128gD. 0.155 g |
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Answer» Correct Answer - B Volume of `O_(2)` at STP`=10mLxx10V=100mL` `22400" mL of " O_(2) at STP =1 mol=4Eq` `100 " mL of " O_(2) STP=(4)/(22400)xx100=(1)/(56)Eq` " Eq of "`KMnO_(4)=" Eq of "O_(2)` `=(1)/(56)" Eq of "KMnO_(4)` `=(1)/(56)xx31.5g of KMnO_(4)` `=0.564g` `(Ew of KMnO_(4)` in acidic medium `=(Mw)/(5)=31.5`) |
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| 244. |
When `SO_(2)` is passed inoto an acidified potassium dichromate soltion, the oxidation number of sulphur and chromium in the final products respectively are:A. `+6,+6`B. `+6,+3`C. `+0,+3`D. `+2,+3` |
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Answer» Correct Answer - B |
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| 245. |
10 g of a mixture of `Cu_2S` and Cus was titrated with 200 " mL of " 0.75 M `MnO_4^(ɵ)` in acidic medium producing `SO_2,Cu^(2+)`, and `Mn^(2+)`. The `SO_2` was boiled off and the excess of `MnO_4^(ɵ)` was titrated with 175 " mL of " `1 M Fe^(2+)` solution. Find the percentage of CuS the in original mixture. |
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Answer» Total `MnO_4^(ɵ)=200xx0.75M MnO_4^(ɵ)` `=150xx5 mEq` of `MnO_4^(ɵ)` `=750 mEq` of `MnO_4^(ɵ)` Excess of `MnO_4^(ɵ)=175 " mL of " M Fe^(2+)` `=175 m mo`l of `Fe^(2+)` `=175 mEq ` of `Fe^(2+)` `=175 mEq` of ` MnO_4^(ɵ)` `MnO_4^(ɵ)` used up `=750-175=575 mEq` `Cu_2to2Cu^(2+)+2e^(-)` `S^(2-)toSO_2+6e^(-)` 1 " mol of "`Cu_2S` required `8e^(-)` `Ew(Cu_(2)S)=159//8` `CuStoCu^(2+)+S^(2-)` `S^(2-)toSO_(2)+6e^(-)` `therefore` Equivalent weight of `CuS=95.5//6` `therefore` x g of `Cu_2S` and `(10-x)g of CuS` react. `therefore((x)/((159)/(8))+(10-x)/((95.5)/(6)))xx1000=575` `thereforex=` Weight of `Cu_2S=4.25g` `therefore` Weight of `CuS=5.75g` % of `CuS=57.5%` |
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| 246. |
In the reaction of canadium oxide (VO) with iron oxide `(Fe_2O_3)` the products are `V_2O_5` and `FeO`. How many grams of `V_2O_5` can be formed from 2.00 g of VO and 5.75 g of `Fe_2O_3`. |
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Answer» `2VO+3Fe_2O_3toV_2O_5+6FeO` Here `Fe_2O_3` is a limiting reagent. `3 " mol of "Fe_2O_3-=1 " mol of "V_2O_5` `(5.75)/(160)=0.036 " mol of "Fe_2O_3=(0.036)/(3)" mol of "V_2O_5` grams `V_2O_5=(0.036)/(3)xx181.9=2.182g` |
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| 247. |
6.0 g of a steel containing sulphur as an impurity was burnt in excess of oxygen, where sulphur is oxidised to `SO_2`. The `SO_2` evolved was oxidised to `SO_4^(2-)` ions by the action of `H_2O_2` solution in the presence of 30 mL solution of 0.04 M NaOH. 25 " mL of " 0.02 M HCl was required to neutralise the excess of NaOH after the above oxidation. Calculate the percentage of sulphur in the given sample of steel (Atomic mass of S is 32). |
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Answer» 6 g steel has S as impurity. `S+O_2toSO_2underset(NaOH)overset(H_2O_2)toSO_4^(2-)` `(10mL,0.04M)` of excess `NaOH-=25mL(0.02M)HCl` `SO_2+H_2O_2+2NaOHtoNa_2SO_4+2H_2O` `1 m mol NaOH-=(1)/(2)mmol SO_2` [Also, 1 m mol `SO_2-=1mmolS`] Excess m" Eq of "`NaOH=m" Eq of "HCl` `=25xx0.02=0.5` `impliesm" Eq of "NaOH used =30xx0.04-0.5` `=1.2-0.5=0.7` `impliesm" mol of "NaOH used =0.7` From stoichiometry: m" mol of "`SO_2=(0.7)/(2)=m" mol of "S` `implies0.35=(weight)/(32)xx1000impliesweight=0.0112g` `%S=(0.0112)/(6)xx100=0.186%` |
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| 248. |
In the reaction `2NH_(3)(g)+5F_(2) to N_(2)F_(4)+6HF` 3.56g `N_(2)F_(4)` is obatained by mixing 2g `NH_(3) and 8g F_(2)`. The percentage yield of the production is: |
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Answer» Correct Answer - a `underset(34g)(2NH_(3)(g))+underset(190g)(5F_(2)) to underset(104g)N_(2)F_(4)+6HF` `"Amount of "N_(2)F_(4)"formed by 2g "NH_(3)=(104)/(34)xx2=6.12g` `"Amount of "N_(2)F_(4)"formed by 8g "F_(2)=(104)/(190)xx8=4.38g` `N_(2)F_(4)` will be limiting and actual amount of the product is 3.56g `"% yield"=("Actual amount of product")/("Calculated amount of product")xx100` `=(3.56)/(4.38)xx100=81.28%` |
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| 249. |
50 " mL of " a mixture of `NH_3` and `H_2` was completely decomposed into `N_2` and `H_2` by sparking. 40 " mL of " `O_2` was then added and the mixture was sparked again. After cooling the mixture was shaken with alkaline pyrogallol and a contraction of 6 mL was observed. Calculate the percentage of `NH_3` in the original mixture. |
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Answer» `underset(x)(NH_3)+underset(y)(H_2)to(1)/(2)underset((x)/(2))(N_2)+(3)/(2)underset((3)/(2)x)(H_2)+underset(ymL)(H_2)` `x+y=50mL` `underset((3)/(2)x)((3)/(2)H_2)+underset((3)/(4)x)((3)/(4)O_2)to(3)/(2)H_2O` `underset(ymL)(H_2)+underset((y)/(2)mL)(1)/(2)O_2tounderset(ymL)(H_2O)`. [After shaki ng with pyrogallol solution, unreacted `O_2` was adsobed and 6 mL contraction was observed Initial volume of `O_2=40ml` unreacted `O_2=6mL`, Reacted `O_2=40-6=34mL`] From equation (ii) and (iii), `O_2` used `(3)/(4)x+(y)/(2)=34implies3x+2y=34xx4` Soving equation (i) and (iv) we get `{:(3x+2y=136),(x+y=50):}]impliesy=14,x=36` Percentage of `NH_3=(36)/(50)=100=72%` Total volume of organic compound `+O_2=95mL` Volume of compound `=20mL` Volume of `O_2` used `=95-20=75mL` |
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| 250. |
`H_2O_2` is reduced rapidly by `Sn^(2+). H_2O_2` is decomposed slowly at room temperature to yield `O_2` and `H_2O_2`. `136g` of `10%` by mass of `H_2O_2` in water is treated with `100mL` of `3M Sn^(2+)` and then a mixture is allowed to stand until no further reaction occurs. The reactions involved are: `2H^(o+)+H_(2)O_(2)+Sn^(2+)toSn^(4+)+2H_(2)` `2H_(2)O_(2)to2H_(2)O+O_(2)` The equivalent of `H_2O_2` reacted with `Sn^(2+)` isA. `0.2`B. `0.3`C. `0.4`D. `0.6` |
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Answer» Correct Answer - D The problem can be solved simply by mol concept since n factor for both `H_(2)O_(2)` and `Sn^(2+)` is 2. `2H^(o+)+underset(1 mol)(H_(2)O_(2))+underset(1mol)(Sn^(2+))toSn^(4+)+2H_(2)O` m" mol of "`H_(2)O_(2)=m" mol of "Sn^(2+)` `=100mLxx3M=300mmol` `=300m" mol of "H_(2)O_(2)` `=300xx2` (n factor) `mEq H_(2)O_(2)` `=600mEq=0.6eq` |
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