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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.
| 451. |
There are two common oxides of sulphur, one of which contains `50% O_(2)` by weight, the other almost exactly `60%`. The weight of sulphur which combine with `1 g` of `O_(2)` (fixed) are in the ratio ofA. (a)`1:1`B. (b)`2:1`C. (c )`2:3`D. (d)`3:2` |
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Answer» Correct Answer - D `{:("First oxide",,,"Second oxide"),("sulphur" 50%,,,40%),("Oxygen" 50%,,,60%):}` In first oxide 1 parts oxygen combines with silphur `=50/50=1` Similarly for second oxide`=40/60=0.67` So the ratio is `1:0.67` or `3:2` |
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| 452. |
Limiting reactant: What mass of water `(H_(2) O)` can be formed by the reaction of `3.00g` of `H_(2) (g)` with `29.0g` of `O_(2)(g)`? Strategy: Using the balanced equation `{:(2H_(2)(g),+,O_(2)(g),,rarr2H_(2)O(l)),(2mol,,1mol,,2mol),(2(2.00g),,32.0g,,2(18.0g)):}` find out the numbers fo moles of each reactatn required to react with the other. Using the given masses, calculate the number of moles of each reactant. Finally, identify the limiting recatant and base the rest of the calculate on it. |
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Answer» Step 1: Calculate the moles of each reactant using the formula, `n = ("Mass")/("Molar mass")` `? "mol" H_(2) = 3.00g H_(2) xx (1 "mol" H_(2))/(2.00g H_(2)) = 1.50 mol H_(2)` `? "mol" O_(2) = 29.0g O_(2) xx (1 "mol" O_(2))/(32.0g O_(2)) = 0.906 "mol" O_(2)` Step 2: Using the balanced equation, determine the required moles of each reactant. The balalnced equation tells us that for every 1 mol `O_(2)` to react completely, 2 mol `H_(2)` is needed. Similarly for every 2 mol `H_(2)` to react completely, 1 mol `O_(2)` is needed, i.e., `(n_(H_(2)))_("required") = 2(n_(O_(2)))_("given")` `(n_(O_(2)))_("required") = (1)/(2) (n_(H_(2)))_("given")` Step 3: Finding the limiting reacant. We see that `2xx(0.906)` or `1.81` mol of `H_(2)` are required to consume `O_(2)` completely but we have just 1.50 mol `H_(2)` , so `H_(2)` is the limiting rectant. Alternatively, we observe that only `(1)/(2) (1.50)` or `0.75` mol `O_(2)` is required to consume `H_(2)` completely but we have 0.906 mol `O_(2)` , so we see again that `H_(2)` is the limiting reacant while `O_(2)` is the excess reacant. Step 4: The reactiion must stop when the limiting reactant, `H_(2)` is used up , so we base the calculation on `H_(2)`. Either we use unitary method or proceed through the mole concept : Unitary method : Accroding to the balanced equation, `4g` of `H_(2)` yields `36g` of `H_(2)O` , thus, `1 g` of `H_(2)` will yield `36//4 g` of `H_(2) O` and `3g` fo `H_(2)` will yield `3xx36//4g = 27g H_(2) O` Mole concept : `underset(H_(2))("g of")rarr underset(H_(2))("mol of")rarr underset(H_(2)O)("mol of")rarr underset(H_(2)O)("g of")` `? g H_(2) O = 3.00g H_(2) = (1 "mol" H_(2))/(2.00g H_(2)) xx (2 "mol" H_(2) O)/(2 "mol" H_(2))` `xx (18.0 g H_(2) O)/(1 "mol" H_(2) O)` `= 27.0g H_(2) O` |
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| 453. |
Limiting reactant: Urea `[(NH_(2))_(2) CO]` used as ferlilzer as animal feed, and in polymer industry, is prepared by the reaction between ammonia and carbon dioxide: `2NH_(3)(g) + CO_(2) (g) rarr (NH_(2))_(2) CO(aq.) + H_(2) O(1)` In one process , `637.2g` of `NH_(3)` is allowed to react with `11.42g` of `CO_(2)` (i) Which of the two reactants is the limiting reactant? (ii) Calculate the mass of `(NH_(2))_(2) CO` formed? (iii) How much of the excess reagent (in grams) is left at the end of the reaction? Strategy: (i) Since we cannot tell by inspection which of the two recantants is the limiting reacant, we have to procced by first converting their masses into number of moles. Take each reactnat in turn and ask how many moles of product (urea) would be obtained if each were completely consumed. The reactant that gives the smaller number of moles of producet is the limiting reactant. (ii) Convert the moles of product obtained to grams of product. (iii) From the moles of product, calculate to grams fo excess reactant needed int he reaction. Then subtract this qunitity from the grams of the reactant available to find the quanity of the excess reactant remaining. |
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Answer» Step 1: Calculate the moles of each reactant. Number of moels `(n) = ("Mass (g)")/("Molar mass" (g mol^(-1)))` `? "mol" NH_(3) = 637.2g NH_(3) xx (1 "mol" NH_(3))/(17.00g NH_(3))` `= 37.48 "mol" NH_(3)` `? "mol" CO_(2) = 11.42g CO_(2) = 1142g CO_(2) xx (1 "mol" CO_(2))/(44.00 g CO_(2))` `= 25.95 "mol" CO_(2)` Step 2 : Convert the moles of each reactant to moles fo product `(NH_(2))_(2) CO`. `37.48 "mol" NH_(3) xx (1 "mol" (NH_(2))_(2) CO)/(2 "mol" NH_(3)) = 18.74 "mol" (NH_(2))_(2) CO` `25.95 "mol" CO_(2) xx (1 "mol" (NH_(2))_(2) CO)/(1 "mol" CO_(2)) = 25.95 "mol" (NH_(2))_(2) CO` Thus, `NH_(3)`, yielding the smaller number of moles fo product must be the limiting reacatant producing `18.74 "mol" (NH_(2))_(2) CO` and `CO_(2)` is the exccss reactant , thus, some `CO_(2)` must be left unconsumed. Step3 : Convert the moles of urea to grams fo urea. `18.74 "mol" (NH_(2))_(2) CO xx (60.00g (NH_(2)) CO)/(1 "mol" (NH_(2))_(2) CO)` `= 1124.4g (NH_(2))_(2) CO` Step 4 : Convert the moles of area to grams of `CO_(2)` (the mass of `CO_(2)` needed to produce this amount of urea) `18.74 "mol" (NH_(2))_(2) CO xx (1 "mol" CO_(2))/(1 mol (NH_(2))_(2) CO) xx (44.00 g CO_(2))/(1 "mol" CO_(2))` `= 824.6 g CO_(2)` Step 5 : Calculate the mass of `CO_(2)` left over. As we started with `11.42 g CO_(2)` the mass of `CO_(2)` remaining un-consumed will be `(1142 g CO_(2)) - (824.63g CO_(2)) = 317.4 g CO_(2)` |
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| 454. |
Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values. `{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}` Using these data, obtain the correct explanation for the following questions. Sodium fusion extract, obtained from anline, on treatment with iron (II) Sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. Hence, the blue colour is due to the formation ofA. `Fe_(4)[Fe(CN)_(6)]_(3)`B. `Fe_(3)[Fe(CN)_(6)]_(2)`C. `Fe_(4) [Fe(CN)_(6)]_(2)`D. `Fe_(3)[Fe(CN)_(6)]_(3)` |
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Answer» Correct Answer - A `Na+C+N rarr NaCN` `Fe^(2+)+6 CN^(-) rarr [Fe(CN)_(6)]^(4-)` In presence of air, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions. `{:(Fe^(2+) rarr Fe^(3+) +e^(-) "]"xx4, E^(@)=-0.77 V),(O_(2)+4 H^(+)+ 4e^(-) + 4e^(-) rarr 2H_(2)O, E^(@)=+1.23 V),(bar(4 Fe^(2+)+4 H^(+)+O_(2) rarr 4 Fe^(3+) +2 H_(2)O", "E_("cell")^(@)= +0.46 V)):}` `Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ion to form ferric ferrocyanide which has prussian blu colour. `4 Fe^(3+)+3[ Fe(CN)_(6)]^(4-) rarr underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))` |
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| 455. |
Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values. `{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}` Using these data, obtain the correct explanation for the following questions. While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution becauseA. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)` |
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Answer» Correct Answer - D Calculate the EMF of all the cell. Only the EMF of the cell involving oxidation of `H_(2)O` to `O_(2)` by `Mn^(3+)` is +ve `{:(Mn^(3+)+e^(-) rarr Mn^(2+)"]"xx4, E^(@)=+1.50 V),(2H_(2)O rarr 4H^(+)+O_(2)+4e^(-), E^(@)=-1.23 V),(bar(4 Mn^(3+)+2 H_(2)O rarr 4 Mn^(2+)+O_(2)+4 H^(+)", "E_("cell")^(@)=+0.27 V)):}` Thus, option (d) is corect. |
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| 456. |
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major idustrial chemicals and is widely used. Nitric is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions `4 NH_(3) (g)+50_(2)(g) overset("Pt/Rh")underset("Catalyst")(rarr) 4 NO(g)+6 H_(2)O(g)` ...(i) `2NO(g)+O_(2)(g) overset(1120 K)(rarr) 2NO_(2)(g)` ...(ii) `3 NO_(2)(g)+H_(2)O(l) rarr 2 HNO_(3)(aq)+NO(g)` ...(iii) In this process the aqueous nitric acid is obtained which can be concentrated by distillation to `~ 68.5%` by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid. If 170 kg of `NH_(3)` is heated in excess of oxygen, then the volume of `H_(2)O(l)` produced in 1st reaction at STP is `(rho_(H_(2)O)=1 g//mL)`A. `33.6xx10^(3) L`B. `270 L`C. `224xx10^(3) L`D. `170 L` |
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Answer» Correct Answer - B Moles of `NH_(3)=170/17=10xx10^(3)` `:.` moles of `H_(2)O` formed `=6/4xx10xx10^(3)` `:.` mass of `H_(2)O` formed `=6/4xx10xx10^(3)xx18` `:.` volume of `H_(2)O` formed `=15xx18L =270 L` |
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| 457. |
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major idustrial chemicals and is widely used. Nitric is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions `4 NH_(3) (g)+50_(2)(g) overset("Pt/Rh")underset("Catalyst")(rarr) 4 NO(g)+6 H_(2)O(g)` ...(i) `2NO(g)+O_(2)(g) overset(1120 K)(rarr) 2NO_(2)(g)` ...(ii) `3 NO_(2)(g)+H_(2)O(l) rarr 2 HNO_(3)(aq)+NO(g)` ...(iii) In this process the aqueous nitric acid is obtained which can be concentrated by distillation to `~ 68.5%` by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid. If 180 L of water completely reacts with `NO_(2)` produced to form nitric acid according to the above reactions, then the volume of air required at STP containing 20% of `NH_(3)` is `(rho_(H_(2)O)=1 g//mL)`A. `1.56xx10^(6) L`B. `6.72 xx10^(4) L`C. `3.36xx10^(6) L`D. None of these |
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Answer» Correct Answer - C Moles of water reacted `=(180xx10^(3))/18=10^(4)` `:.` moles of `NO_(2)` required `=30xx10^(3)=` mole of `NH_(3)` required `:.` volume of air at `STP=(30xx10^(3)xx22.4)/0.2=3.36xx10^(6) L`. |
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| 458. |
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major idustrial chemicals and is widely used. Nitric is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions `4 NH_(3) (g)+50_(2)(g) overset("Pt/Rh")underset("Catalyst")(rarr) 4 NO(g)+6 H_(2)O(g)` ...(i) `2NO(g)+O_(2)(g) overset(1120 K)(rarr) 2NO_(2)(g)` ...(ii) `3 NO_(2)(g)+H_(2)O(l) rarr 2 HNO_(3)(aq)+NO(g)` ...(iii) In this process the aqueous nitric acid is obtained which can be concentrated by distillation to `~ 68.5%` by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid. 85 kg of `NH_(3) (g)` was heated with 320 kg oxygen in the first step and `HNO_(3)` is prepared according to the above reactions. If the final solution has volume 500 lt. Then molarity of `HNO_(3)` isA. 3.33 MB. 8 MC. 2 MD. 6.66 M |
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Answer» Correct Answer - D `4 NH_(3)(g)+5O_(2)(g) rarr 4NO+6H_(2)O` `5xx10^(3)` mole `10^(4)` mole `5xx10^(3)` `{:(2NO(g)+O_(2)(g) rarr 2NO_(2) (g)),(5xx10^(3)" "5xx10^(3)),(3NO_(2)+H_(2)O(l) rarr 2HNO_(3)(g)+NO(g)),(5xx10^(3)" "2/3xx5xx10^(3)):}` Molarity `=2/3xx(5xx10^(3))/500=6.66 M`. |
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| 459. |
If 6.3 g of NaHC`O_(3)` are added to 15.0 g `CH_(3)`COOH solution, the residue is found of weight 18.0 g. What is the mass of `CO_(2)` released in the reaction?A. 4.5 gB. 3.3gC. 2.6gD. 2.8g |
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Answer» b) According to law of conservation of mass, Mass of reactants = Mass of products `therefore` 6.3 + 15.0 = 18.0 + x Or x = 21.3 -18.0 = 3.3 g |
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| 460. |
Density of an object is `g//cm^(3)`. Express it in `kg//m^(3)`. |
| Answer» Correct Answer - `3000 kg m^(-3)` | |
| 461. |
Strictly speaking the term formula mass has no meaning forA. radon gasB. nitrogen dioxide gasC. ordinary table saltD. sugar |
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Answer» Correct Answer - A It Is a nobel gas composed of isolates atoms, i.e., monoatomic molecules. |
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| 462. |
Ammonia gas is passed into water, yielding a solution of density `0.93 g//cm^(3)` and containing `18.6% NH_(3)` by weight. The mass of NH3 per cc of the solution is :A. 0.17 g / `cm^(3)`B. 0.34 g / `cm^(3)`C. 0.51 g / `cm^(3)`D. 0.68 g / `cm^(3)` |
| Answer» Correct Answer - A | |
| 463. |
The mass of 11.2 L of ammonia gas at STP isA. `8.5` gB. 85 gC. 17 gD. `4.25` g |
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Answer» Correct Answer - A Molar mass of `NH_(3) = 14 + 3 xx 1 = 17 g mol^(-1)` `therefore` Mass of `22. 4 ` L of `NH_(3)` at STP = 17 g `therefore` Mass of `11.2 ` L of ` NH_(3) ` at STP = `8.5` g |
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| 464. |
What quantity of copper(II) oxide will react 2.80litre of hydrogen at NTPA. 79.5 gB. 2 gC. 9.9 gD. 22.4 g |
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Answer» `underset(79.5g)(CuO)+underset(22.4L)(H_2) to Cu + H_2O` 22.4 L of `H_2 -=` 79.5g of CuO 2.80 L of `H_2 -=(79.5)/(22.4)xx2.80=9.9g` of CuCO |
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| 465. |
The measured density at `NTP` of `He` is `0.1784 g L^(-1)`. Calculate the weight of `1 "mole"` of `He`.A. 39.9 gB. 22.4 gC. 3.56 gD. 29 g |
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Answer» 1 mole occupies a volume of 22.4 L at STP Mass of 1 mole of gas = Density x Volume `=1.78xx22.4=39.9g` |
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| 466. |
1 mole of methylamine on reaction with nitrous acid gives at NTP:A. 1.0 Litre of nitrogenB. 22.4 Litre of nitrogenC. 11.2 Litre of nitrogenD. 5.6 Litre of nitrogen |
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Answer» Correct Answer - B Methyl amine reacts with nitrous acid to form methyl alcohol and nitrogen is evolved. `CH_(3)NH_(2)+HONO rarr CH_(3)OH+N_(2)+H_(2)O` 1 mol of methyl amine on reaction with `HNO_(2)` gives 22.4 litre of `N_(2)` at N.T.P. |
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| 467. |
A compound contains `3.2%` of oxygen. The minimum mol. Wt. of the compound isA. (a)`300`B. (b)`440`C. (c )`350`D. (d)`500` |
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Answer» Correct Answer - d The compound must contain at least one oxygen atom So, a minimum of 1 g atom of oxygen will be present in 1 g molecule, i.e., 1 mole of the compound. If M is the mol. Wt. of the compound then since 16 is the atomic mass of oxygen so minimum of `16 g` of oxygen will be present in `M g` of the compound Thus, `%` of oxygen `=16/Mxx100` or `3.2=(16xx100)/M` or `M=500` |
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| 468. |
Define science in short. |
| Answer» Science is a continuting human effort to systematize knowledge for describing and understanding nature. | |
| 469. |
The correctly reported difference of 19.3226 and 8.06 will have siginificant figures equal toA. ThreeB. FourC. SixD. Five |
| Answer» Correct Answer - B | |
| 470. |
How many moles of sodium chloride present in 250 mL of a 0.50 M NaCl solution ?A. 0.125 molB. 0.150 molC. 0.075 molD. 0.02 mol |
| Answer» Correct Answer - A | |
| 471. |
The oxide of an element contains `67.67%` of oxygen. Equivalent weight of the element isA. `2.46`B. `3.82`C. `4.36`D. `4.96` |
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Answer» Correct Answer - B Oxygen `= 67.67 `g Metal `= 32.33 ` g `67.67` g oxygen `equiv 32.33` metal 8 g oxygen `=(32.33)/(67.67)xx8.0=3.82` g |
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| 472. |
The equivalent weight of a certain trivalent element is 20. Molecular weight of its oxide isA. `168`B. `68`C. `152`D. `56` |
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Answer» Correct Answer - A At mass of element `= 20xx3=60` Formula of its oxide is `M_(2)O_(3)` `:.` Molecular mass of its oxide `= 2xx60+16xx3` `= 120 + 48 = 168` |
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| 473. |
A metal oxide contains 60% metal . The equivalent weight of metal isA. 12B. 60C. 40D. 24 |
| Answer» Correct Answer - A | |
| 474. |
The mole fraction of glucose in aqueous solution is 0.2 then molality of solution will beA. 13.8B. 55.56C. 2D. 12 |
| Answer» Correct Answer - A | |
| 475. |
Dissolving `120g` of urea `(Mw = 60)` in `1000 g` of water gave a solution of density `1.15 g mL^(-1)`. The molarity of solution is:A. `1.78 M`B. `1.02 M`C. `0.50 M`D. `2.05 M` |
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Answer» Correct Answer - D `n_("urea") = ("Mass"_("urea"))/("Molar mass"_("urea")) = (120g)/(60g mol^(-1)) = 2 "mol"` Volume of solution `= ("Mass of solution")/("Density of solution")` `= ((120+1000)g)/(1.15g mL^(-1)) = 974 mL = 0.974 L` Molarity `= ("Moles of urea")/("Liters of solution") = (2 mol)/(0.974L)` `= 2.05 "mol" L^(-1) = 2.05M` |
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| 476. |
Normally: An aquous solution contains `4.202 g` of `HNO_(3)` in `600 mL` of solution. Calculate the normally of solution? Stragegy: Convert grams of `HNO_(3)` to moles of `HNO_(3)` to moles of `HNO_(3)` and then to equivalents of `HNO_(3)`. Finally , apply the defintion of normally. `(g HNO_(3))/(L) rarr (mol HNO_(3))/(L) rarr (eq HNO_(3))/(L) rarr (eq HNO_(3))/(L) = N HNO_(3)` |
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Answer» Step 1: Calculate the number of moles. mol `HNO_(3) = ("Grams" HNO_(3))/("Molar mass" HNO_(3)) = (4.202g)/(63.02g mol^(-1))` `= 0.0666 mol HNO_(3)` Step 2: Calculate the number of equivalnets of `HNO_(3)`. Number of equivalnents of `HNO_(3)` (No. eq `HNO_(3)`) = (Number of moles of acid) `xx` (Number of hydrogen ions furnished) `= (0.0666 mol HNO_(3))(1)` `= 0.0666 eq HNO_(3)` Step 3: Calcularte normally. Normally `(N) = ("No. eq" HNO_(3))?("Liners of solution")` `= (0.0666 eq HNO_(3))/(0.600 L)` `= 0.11 NHNO_(3)` Alterntively using single setup, `? (eq HNO_(3))/(L) = (4.202 g HN O_(3))/(0.600 L) xx (1 mol HNO_(3))/(63.02g HNO_(3))` `xx (1 eq HNO_(3))/(1 mol HNO_(3))` `= 0.0111 N NHO_(3)` |
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| 477. |
One torr is equal toA. 1 atm. PressureB. 1 mm of HgC. 1 cm of HgD. 1 m of Hg |
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Answer» Correct Answer - B 1 torr = 1 mm of Hg |
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| 478. |
Boron has two stable isotopes, `.^(10)B(19%)` and `.^(11)B(81%)`. The atomic mass that should appear for boron in the periodic table isA. (a)`10.8`B. (b)`10.2`C. (c )`11.2`D. (d)`10.0` |
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Answer» Correct Answer - A Atomic mass`=(10xx19+81xx11)/100` `=(190+891)/100` `=1081/100=10.81` |
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| 479. |
One fermi isA. `10^(-13)` cmB. `10^(-15)` cmC. `10^(-10)` cmD. `10^(-12)` cm |
| Answer» Correct Answer - A | |
| 480. |
One fermi isA. (a)`10^(-13) m `B. (b)` 10^(-15) m`C. (c )` 10^(-10) m`D. (d)` 10^(-12) m` |
| Answer» Correct Answer - A | |
| 481. |
Fermi is a unit of length for measuring the nuclear diameter. It is equal toA. `10^(-10) m`B. `10^(-13) m`C. `10^(-15) m`D. `10^(-12) m` |
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Answer» Correct Answer - C 1 Fermi = `10^(-15) m` |
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| 482. |
One litre hard water contains 12.00 mg `Mg^(2+)` millieqivalent of washing soda required to remove its hardness isA. (a)1B. (b)12.15C. (c )`1xx10^(-3)`D. (d)`12.5xx10^(-3)` |
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Answer» Correct Answer - A Meq of `Mg^(+2)`= Meq of washing soda `W/Exx1000=Mg^(+2), EW=24/2=12` `(12xx10^(-3))/12xx1000=1` |
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| 483. |
In which of the following numbers all zeros are significant ?A. 0.0007B. 0.007C. 70D. 0.7 |
| Answer» Correct Answer - C | |
| 484. |
In which of the following numbers all zeros are significant ?A. `0*0005`B. `0*0500`C. `50*000`D. `0*00050` |
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Answer» Correct Answer - C In `50*000` all zeros are significant |
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| 485. |
In which of the following numbers are all the zeros significant?A. `10.00`B. `0.200`C. `0.0020`D. `0.00002` |
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Answer» Correct Answer - A Zeros at the end of a number (greater than 1) that contains a decimal point are always significiant. |
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| 486. |
Assertion: A one mola solution prepared at `20^(@)`C will retain the same molality at `100^(@)`C, provided there is no loss of solute or solvent on heating. Reason: Molality is independent of temperature.A. (a)If both assertion and reason are true and the reason is the correct explanation of the assertion.B. (b)If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. (c )If assertion is true but reason is false.D. (d)If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 487. |
`100 g CaCO_(3)` reacts with `1 litre 1 N HCl`. On completion of reaction how much weight of `CO_(2)` will be obtainA. (a)`5.5 g `B. (b)` 11 g`C. (c )`22 g `D. (d)` 33 g` |
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Answer» Correct Answer - C `underset(100g)(CaCO_(3))+underset(2N)(2HCl) rarr CaCl_(2)+underset(44g)(CO_(2))+H_(2)O` `100 g CaCO_(3) with 2 N HCl gives 44 g O_(2)` `100 g CaCO_(3) with 1 N HCl gives 22g O_(2)` |
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| 488. |
A given solution of NaOH contains 2.00 g of NaOH per litre of solution. Calculate the molarity of this solution. |
| Answer» Correct Answer - `0.05 M` | |
| 489. |
Calculate the normality of solution of 4.9 % (w/v) of `H_(2)SO_(4)`. |
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Answer» Weight of solute = 4.9 g Volume of solution = 100 ml Equivalent weight of `H_(2)SO_(4) = 49` `N=(4.9)/(49)xx(1000)/(100) = 1`. |
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| 490. |
A compound having the formula `Br_(3) C_(6)H_(3) (C_(3) H_(g))_(n)` contains `10.46% Br` by mass, the value of `n` isA. `54`B. `65`C. `45`D. `35` |
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Answer» Correct Answer - C `% Br = (3 xx "AM of Br")/("MM of compound") xx 100%` (AM is atomic mass, MM is molar mass) `10.46% = (3(79.9u))/(314.7 + 44n) xx 100%` `n = 45` |
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| 491. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If correct answers to the question number A,B,C and D (say) are 4,0,9 and 2 respectively, then correct darkening of bubbles should be as shown on the side. (C) Reaction of `Br_(2)` with `Na_(2)CO_(3)` in aqueous solution gives sodium bromide and sodium bromate with evolution of `CO_(2)` gas. The number of sodium bromide molecules involved in the balanced chemical equation is ..................... |
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Answer» Correct Answer - 5 `3Br_(2)+3Na_(2)CO_(3)to5NaBr+NaBrO_(3)+3CO_(2)` |
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| 492. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If correct answers to the question number A,B,C and D (say) are 4,0,9 and 2 respectively, then correct darkening of bubbles should be as shown on the side. (B) A student performs a titration with different buretters and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value is ........ |
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Answer» Correct Answer - 3 The least precise terms in the litre value, i.e., 25.2 and 250 has 3 significant figures. |
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| 493. |
The density (in g `mL^(-1)`) of a 3.60 M sulphuric acid solution that is 29% of acid by mass isA. `1.45`B. `1.64`C. `1.88`D. `1.22` |
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Answer» Correct Answer - D 3.6 M `H_(2)SO_(4)` means `3.6xx98=352.8g` of sulphuric acid are present in 1000 mL of the solution 29.0 g of acid are present in solution = 100 g 352.8 g of acid are present in solution `=((100g)xx(352.8g))/((29g))=1216g` Density of solution `= ("Mass of solution")/("Volume of solution")` `=(1216g)/(1000mL)=1.22gmL^(-1)` |
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| 494. |
Volume occupied by one molecule of water (density = 1 g `cm^(3)`)A. `3.0xx10^(-23) cm^(3)`B. `5.5xx10^(-23)cm^(3)`C. `9.0xx10^(-23)cm^(3)`D. `6.023xx10^(-23)cm^(3)` |
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Answer» Correct Answer - A Mass of 1 molecule of water `=(18)/(6.022xx10^(23))=3xx10^(-23)g`. Density of water = 1.00 g `cm^(-3)` `:.` volume occupied by one molecule of water `=3xx10^(-23)cm^(-3)` |
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| 495. |
Which one of the following has the highest oxidation number of iodine?A. `KI_(3)`B. `KI`C. `IF_(5)`D. `KIO_(4)` |
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Answer» Correct Answer - D `Koverset(**)(I)O_(4)` `1+x-2xx4=0, x=8-1=+7` |
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| 496. |
Which one of the following is the highest?A. 0.2 mole of hydrogen gasB. `6.023xx10^(22)` molecules of nitrogenC. 0.1 g of silverD. 0.1 g mole of oxygen |
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Answer» Correct Answer - C (A) 0.2 mol of `H_(2)=0.2xx2=0.4g` (B) `6.023xx10^(23)` molecules of `N_(2)=1` mol =28 g (C) 0.1 g (D) 0.1 mole of `O_(2)` is the highest |
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| 497. |
The total number of electrons present in `18 mL` of water is ……A. `6.023xx10^(25)`B. `6.023xx 10^(24)`C. `6.023xx18xx10^(23)`D. `6.023xx10^(23)` |
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Answer» Correct Answer - B 18 mL `H_(2)O = 18 g H_(2) O` `=18/18 = 1 ` mol `H_(2)O ` `=6. 0 23 xx 10^(23)` molecules electron in 1 mol = 10 `therefore` Total electrons in 18 mL `H_(2) O = 6. 0 2 xx 10 ^(23) xx 10` `= 6. 02 xx10^(24)` |
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| 498. |
20.0 kg of `N_(2)(g)` and 3.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. The amount of `NH_(3)(g)` formed isA. 17 kgB. 34 kgC. 20 kgD. 3 kg |
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Answer» Correct Answer - A `underset(28" g")(N_(2)) + underset(6" g")(3H_(2)) rarr underset(34" g")(2NH_(3))` Limiting reactant is `H_(2)` Weight of `NH_(3)` obtained from `6xx 10^(-3) ` kg of `H_(2)` `= 34xx 10^(-3) ` kg `therefore` Weight of `NH_(3)` obtained from 3 kh of `H_(2)` `= (34xx 10^(-3))/(6xx 10^(-3)) xx 3 ` = 17 kg `NH_(3)` |
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| 499. |
`O_(2)^(2+)` is the symbol of ...... IonA. oxideB. superoxideC. peroxideD. monoxide |
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Answer» Correct Answer - C `O_(2)^(2-)` is the symbol of peroxide |
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| 500. |
In a balanced equation `H_(2)SO_(4)toxHItoH_(2)S+yI_(2)+zH_(2)O` The values of x,y,z areA. `x =3, y=5, z=2`B. `x=4, y=8, z=5`C. `x=8, y=4, z=4`D. `x=5, y=3, z=4` |
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Answer» Correct Answer - C `H_(2)SO_(4) + HI to H_(2)S+I_(2)+H_(2)O` Balancing by partial equation method, `({:(,,2HI,to,2[H],+,I_(2)xx4),(H_(2)SO_(4),+,8[H],to,4H_(2)O,+,H_(2)S):}/((H_(2)SO_(4),+,8HI,to,4H_(2)O,+,4I_(2)+S)))/` c.f., `H_(2)SO_(4) + xHI to H_(2)S + yI_(2)+zH_(2)O` `:. x = 8, y = 4, z = 4` |
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