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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 451. |
Equal weights of methane and oxygen are mixed in an empty container at `25^(@)C`. The fractin of the total pressure exerted by oxygen isA. `(1)/(3)`B. `(1)/(2)`C. `(2)/(3)`D. `(1)/(3) xx (273)/(298)` |
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Answer» Correct Answer - A Let `32 g` of each be present. Moles of `O_(2) = (32)/(32) = 1` Moles of `CH_(4) = (32)/(16) = 2` Mole fraction of `O_(2) = (1)/(1 + 2) = (1)/(3)` which is same as the fraction of pressure. |
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| 452. |
The temperature at which molarity of pure water is equal to its molality is `:`A. `273K`B. `298K`C. `277K`D. None |
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Answer» Correct Answer - 3 At `4^(@)C` density of water is `1 g//ml.` |
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| 453. |
Why is limiting reactant so named ? |
| Answer» Limiting reactant i.e, the reactant present in fixed amount is so because it limits the participation of other reactants even if present in excess in particular reaction. | |
| 454. |
`K_(2)Cr_(2)O_(7)+C_(2)O_(4)^(2-)+H_(2)SO_(4)rarr K_(2)SO_(4)+CO_(2)+Cr_(2)(SO_(4))_(3)+H_(2)O` In above reaction, identify the elements which do not undergo change in their oxidation state `:`A. `C`B. `S & C`C. `K,O,S & H`D. `C & O` |
| Answer» Correct Answer - 3 | |
| 455. |
In the balanced equation `FeS+MnO_(4)^(-)overset(H^(+))rarrFe^(3+)+SO_(4)^(2-)+Mn^(2+)` the stoichiometric coefficients of `FeS` and `MnO_(4)^(-)` are in the ratio `:`A. `8:5`B. `5:8`C. `9:5`D. `5:9` |
| Answer» Correct Answer - 4 | |
| 456. |
The modern atomic mass unit if based on the mass of ……… |
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Answer» Correct Answer - A::B::C The modern atomic mass unit is based on the mass of `C - 12`. |
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| 457. |
Which of the following `is//are` correct? The following reaction occurs: `Na_(2) CO_(3) + 2HCl rarr 2NACl + CO_(2) + H_(2) O` `106.g "of" Na_(2) CO_(3)` reacts with `109.5 g "of" HCl`.A. The `HCl` is in excess.B. `117.0 g` of `NaCl` is formed.C. The volume of `CO_(2)` produced at 1 bar and `273 K` is `22.7 L`D. The volume of `CO_(2)` produced at 1 bar and `298 K` is `24.7 L` |
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Answer» Correct Answer - A::B::C::D `(Mw "of" Na_(2) CO_(3) = 106 "of" HCl = 36.5, Mw "of" NaCl = 58.5)` Mole of `Na_(2) CO_(3) = (106)/(106) = 1.0 "mol"` Moles of `HCl = (109.5)/(36.5) = 3.0 "mol"` a. Since for a mol of `Na_(2) CO_(3)`, 2 mol of `HCl` is required. So, `HCl` is in excess `(3 - 2) = 1.0 "mol"` Therefore, `Na_(2) CO_(3)` is the limiting quaintity. b. weight of `NaCl` formed `= (1.0 "mol" Na_(2) CO_(3)) ((2 "mol" NaCl)/("mol" Na_(2) CO_(3)))((58.8 g NaCl)/("mol" NaCl))` 1 mol of `Na_(2) CO_(3) = 1` mol of `CO_(2) = 22.7 L` at 1 bar, `273 K` 1 mol of `Na_(2) CO_(3) = 1` mol of `CO_(2) = 24.7 L` at 1 bar, `298 K` |
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| 458. |
Salt cake `(Na_(2)SO_(4))` is prepared as follows: `2NaCl + H_(2) SO_(4) rarr Na_(2)SO_(4) + 2HCl` How much 80% pure salt cake could be produced form `100.0 g` of 90% pure salt in the above reaction?A. `43.92 g`B. `68.62 g`C. `87.84 g`D. `137.25 g` |
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Answer» Correct Answer - D Weight of `Na_(2) SO_(4) = (109.8 g Na_(2) SO_(4)) xx` `((100 g "new mixture")/(80.0 g Na_(2)SO_(4)))` `= (109.8 xx 100)/(80)` `= 137.25 g` mixture |
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| 459. |
Silver (atomic weight `108 g mol^(-1))` has a density of `10.5 g cm^(-3)`. The number of silver atoms on a surfaces of area `10^(-12) m^(2)` can be expressed in scientific notation as `Y xx 10^(-x)`, The value of `x` is ……. |
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Answer» Density `= 10.5 g c c^(-1)` This measn `10.5 g` silver is present in `1 cm^(3)`. `(10.5)/(108)` mol silver is present in `1 cm^(3)`. `(10.5)/(108) N` silver atoms present in `1 cm^(3)` `.^(3)sqrt((10.5)/(108) N)` silver atoms present in `1 cm` `(.^(3)sqrt((10.5)/(108) N))^(1//2)` silver atoms present is `1 cm^(2)`. `10^(-12)` is equal to `10^(-8) cm^(2)`. `(.^(3)sqrt((10.5)/(108) N))^(1//2) xx 10^(-8)` silver atoms present in `10^(-12) m^(2)`. On solving, we get `1.5 xx 10^(-7) = Y xx 10^(-x)` `:.x = 7` |
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| 460. |
Which of the following `is//are` correct. The following reaction occurs: ltrgt `CS_(2) + 3Cl_(2) overset(Delta)rarr C Cl_(4) + S_(2) Cl_(2)` `1.0 g` of `CS_(2)` and `2.0 g` of `Cl_(2)` reacts.A. `0.714 g CS_(2)` is used in the reaction.B. `0.286 g CS_(2)` is in formed.C. `1.45 g of C Cl_(4)` is formedD. `0.8 g Cl_(2)` is in excess |
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Answer» Correct Answer - A::B::C `(Mw "of" CS_(2) = 76, Mw "of" Cl_(2) = 71, Mw "of" C Cl_(4) = 154 g mol^(-1))` Weight of `Cl_(2)` needed `=(1.0 g CS_(2)) ((1 "mol" CS_(2))/(76 g CS_(2)))((3 "mol" CS_(2))/("mol"CS_(2)))((71 g Cl_(2))/("mol" Cl_(2)))` `= (1 xx 3 xx 71)/(76) = 2.8 g Cl_(2)` needed Since there is `2.0 g Cl_(2)` is the limiting quantity a. Weight of `Cs_(2)` used `= (2.0 g Cl_(2))((1 "mol" Cl_(2))/(71 g Cl_(2))) ((1 "mol" CS_(2))/(3 "mol" Cl_(2)))((76 g CS_(2))/("mol" CS_(2)))` `= (2 xx 1 xx 1 xx 76)/(71 xx 3) = 0.714 g CS_(2)` used. b. Weight of `CS_(2)` excess or formed `= (1.0 g CS_(2)` present) - `(0.714 g` used) `= 0.286 g CS_(2)` formed c. Weight of `C Cl_(4)` formed `= (2.0 g Cl_(2)) ((1 "mol" Cl_(2))/(71 g Cl_(2)))((1"mol" CCl_(4))/(2"mol"Cl_(2)))` `((154 g C Cl_(4))/("mol" C Cl_(4)))` `= (2 xx 1 xx 154)/(71 xx 3) = 1.45 g C Cl_(4)` d. Wrong |
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| 461. |
Among the following, what is the number of elements showing only one non-zero oxidation state? |
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Answer» Correct Answer - B `O,Cl,underline(F),N,P,Sn,Tl,underline(Na),Ti` `Na` shows only + 1 `F` shows only - 1 |
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| 462. |
The total number of electrons present in `18 mL` of water is …… |
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Answer» Correct Answer - A::B::C::D The total numbe of electrons present in `18 mL` of water is `6.023 xx 10^(24)`. Number of electron in one molecules of `H_(2) O` is `2 + 8 = 10` Denisty = 1 `18 mL` means `18 g` Moles `= (18)/(18) = 1` Moles `= 6.023 xx 10^(23)` Electrons `= 6.023 xx 10^(23) xx 10 = 6.023 xx 10^(24)` |
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| 463. |
`2.66 g` of chloride of a metal when treated with silver nitrate solution, gave `2.87 g` of silver chloride `3.37 g` of another chloride of the same metal gave `5.74 g` of silver chloride when treated with silver nitrate solution. Show that the results illustrate the law of multiple proportions. |
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Answer» `AgNO_(3)+MClrarrMNO_(3)+underset(("white ppt"))(AgCl)` Molecular mass of `AgCl=(108+35.5)=143.5g` In the first sample `143.5 g` of `AgCl` contain `Cl = 35.5 g` `2.87 g` of `AgCl` contain `Cl=((35.5g))/((143.5g))xx(2.87g)=0.71g` Now, chlorine `(Cl)` present in `AgCl` has actually been obtained from metal chloride `:.` Mass of metal (M) in the sample `= (2.66-0.71) = 1.95 g` In the second sample `143.5 g` of `AgCl` contain `Cl = 35.5 g` `5.74 g` of `AgCl` contain `Cl=((35.5g))/((143.5g))xx(5.74g)=0.42g` Mass of metal (M) in the sample `= (3.37 - 1.42) = 1.95 g` From the above data, it is clear that in both the samples of metal chlorides, mass of metal `(M) = 1.95 g` `:.` Ratios of the masses of chlorine `(Cl)` which combine with a fixed mass of metal `= 0.71 : 1.42 or 1:2`. As the ratio is a simple whole number ratio therefore, the law of multiple proportions is illustrated. |
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| 464. |
Which of the following statements `is//are` correct" The following reaction occurs: `2Al + 3MnO overset(Delta)rarr Al_(2) O_(3) + 3 Mn`. `108.0 g` of `Al` and `213.0 g` of `MnO` was heated to initiate the reaction. `(Mw "of" MnO = 71`, atomic weight of `Al = 13)`A. `Al` is present in excessB. `MnO` is present is excess.C. `54.0 g` of `Al` is requiredD. `159.0 g` of `MnO` is in excess. |
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Answer» Correct Answer - A::C Moles of `Al = (108)/(27) = 4.0 "mol"` moles of `MnO = (213)/(71) = 3.0 "mol"` a. Since 2 mol of `Al` requires 3 mol of `MnO`, therefore `Al` is in excess. b. Wrong Weight of `Al` required `= (3.0 "mol" MnO) ((2 "mol" Al)/(2 "mol" MnO))((27 g Al)/("mol" Al))` `= (3 xx 2 xx 27)/(3) = 54.0 g` of `Al` d. Weight of `Al` in excess `= (108 - 54) = 54.0 g` |
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| 465. |
In the reaction: `2Al + Cr_(2) O_(3) rarr Al_(2) O_(3) + 2 Cr`, `4.98 g` of `Al` reacted with `20.0 g Cr_(2)O_(3)`. How much grams of reactant ramains at the completion of the reaction? |
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Answer» (Atomic weight of `Al` and `Cr = 27` and 52, `Mw` of `Cr_(2)O_(3) = 152)` Moles of `Al = (4.98)/(27 Al) = 0.184 "mol"` `= (0.184)/(2) = 0.92 "mol of" Cr_(2) O_(3)` Since 2 mol `Al` is required for 1 mol of `Cr_(2)O_(3)`. So, `Al` is the limiting reagent and `Cr_(2)O_(3)` is excess. Moles of `Cr_(2)O_(3)` in excess `= (0.131 - 0.092) = 0.039` `~~ 0.04 "moo"` Weight of `Cr_(2) O_(3)` is excess `= 0.04 xx 150 ~~ 6 g Cr_(2) O_(3)` |
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| 466. |
A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm. The mole fraction of `C_(2)H_(4)` in the mixture isA. 0.25B. 0.75C. 0.45D. 0.55 |
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Answer» Correct Answer - B `chi_(C_(2)H_(4))` or `chi_(b) = 1 - 0.25 = 0.75` |
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| 467. |
What is the percentage of aluminium in `Al_(2)O_(3)`? `(Al = 27, O = 16)` |
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Answer» `Mw of Al_(2) O_(3) = 27 xx 2 + 16 xx 3 = 102 g` `% of Al = (54 xx 100)/(102) = 52.94%` |
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| 468. |
Suppose elements `X` and `Y` combine to form two compounds `XY_(2)` and `X_(3)Y_(2)` when 0.1 mole of former weigh `10 g` while 0.05 mole of the latter weigh `9g`. What are the atomc weights of `X` and `Y`.A. 40,30B. 60,40C. 20,30D. 30,20 |
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Answer» Correct Answer - A `X + 2Y rarr XY_(2)` 0.1 mol of `XY_(2) = 10 g` 1 mol of `XY_(2) rarr 100 g` ii. `3X + 2 Y rarr X_(3) Y_(2)` 0.05 mol of `X_(3)Y_(2) = (9)/(0.05) = 180 g` ` {:( :. X + 2 Y = 100),(3X + 2Y = 180):}] {:(MwXY_(2) = 100),(MwX_(2)Y_(3) = 180):}` Solve of `X` and `Y` `X = 40 g` `Y = 30 g` |
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| 469. |
Equal weight of Aluminium and Oxygen are allowed to combine with each other to produce `Al_2O_3`.Identify the correct statement. `(4Al+3O_2to2Al_2O_3)`(At. Wt. of Al=27, O=16)A. Aluminium metal is the limiting reagentB. The fraction of excess reagent left unreacted is `1/9`C. The mass of `Al_2O_3` produced is `51/27` times the mass of oxygen taken initially.D. The mass of aluminium and oxygen is left unreacted and the mass of `Al_2O_3` produced is double the mass of aluminium taken initially. |
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Answer» Correct Answer - A::B::C |
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| 470. |
If 1/2 moles of oxygen combine with aluminium to form `Al_(2)O_(3)` then weight of Aluminium metal used in the reaction is (Al = 27 )A. 27 gB. 18 gC. 54 gD. 40.5 g |
| Answer» Correct Answer - B | |
| 471. |
The molarity of `H_(2)SO_(4)` is `18 M`. Its density is `1.8 g mL^(-1)`.A. 36B. 200C. 500D. 18 |
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Answer» Correct Answer - C Use formula `d = M ((Mw_(2))/(1000) + (1)/(m)) implies 1.8 = 18 ((98)/(1000) + (1)/(m))` solve for `m:` `:. M = 500` |
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| 472. |
Molarity of `H_(2)SO_(4)` is `18 M`. Its density is `1.8 g//cm^(3)`, hence molaity is :A. `18`B. `100`C. `36`D. `500` |
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Answer» Correct Answer - D molarity `(Mxx1000)/(1000d-MM_(w))=(18xx1000)/(1000xx1.8-18xx98)=500` |
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| 473. |
If `1 1/2` moles of oxygen combine with Al to form `Al_(2)O_(3)` the weight of Al used in the reaction is (Al=27)A. `27 g`B. `54 g`C. `40.5 g`D. `81 g` |
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Answer» Correct Answer - B `{:(3O_(2)+4Alrarr2Al_(2)O_(3)O),(3rarr4),(1rarr(4)/(3)xx(3)/(2)),((3)/(2)rarr=2molxx27=54 g):}` |
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| 474. |
If `1 1/2` moles of oxygen combine with Al to form `Al_(2)O_(3)` the weight of Al used in the reaction is (Al=27)A. 27gB. 54gC. 40.5gD. 81g |
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Answer» Correct Answer - B |
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| 475. |
A solution contains 4 g of NaOH and 16.2 g of water. The mole fraction of solute and solvent are respectively.A. 0.1,0.9B. 0.2,0.8C. 0.5,0.5D. 0.4,0.6 |
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Answer» Correct Answer - A `n_(B)=((4g))/(("40 g mol"^(-1)))=0.1` mol `n_(A)=(("16.3g"))/(("18 g mol"^(-1)))=0.9` mol `X_(B)=(n_(B))/(n_(B)+n_(A))=(0.1)/((0.1+0.9))=0.1` mol `X_(A)=1-0.1=0.9`. |
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| 476. |
A compound contains 28% N and 72% of a metal by weight . Three atoms of metal combine with two atoms of N . Find the atomic weight of metal. |
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Answer» Correct Answer - 24 |
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| 477. |
Which has maximum number of molecules among the following ?A. 44 g of `CO_(2)`B. 48 g of `O_(3)`C. 8 g of `H_(2)`D. 64 g of `SO_(2)` |
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Answer» Correct Answer - C No. of moles of `CO_(2)=((44g))/(("44 g mol"^(-1)))=1` mol No. of moles of `O_(3)=((48g))/(("48 g mol"^(-1)))=1` mol No. of moles of `H_(2)=(8g)/(("2g mol"^(-1)))=4` mol No. of moles of `SO_(2)=((64g))/(("64g mol"^(-1)))=1` mol `:.` 4 moles or 8g of `H_(2)` have max. no of molecules. |
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| 478. |
If 1.5 moles of oxygen combine with Al to form `Al_(2)O_(3)`. The mass of Al in grams. (Atomic mass of Al = 27) used in the reaction is :A. 2.7B. 54C. 40.5D. 81 |
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Answer» Correct Answer - B `underset("2 mol")(2Al+)3//2underset("1.5 mol")(O_(2)rarr)underset("1 mol")(Al_(2)O_(3))` Mass of Al in grams used `=2xx27=54` g |
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| 479. |
Methyl-t-butel ether, `C_5H_12O` is added to gasoline to promote cleaner burning How many moles of oxygen gas. `O_2` are required to burn 1.0 mol of this compound completeley to form carbon dioxide and water ?A. 4.5 molB. 6.0 molC. 7.5 molD. 8.0 mol |
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Answer» Correct Answer - C |
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| 480. |
Calcualate the molarity and molality of 20% aqueous ehtanol `(C_(5) H_(5) OH)` solution by volume. (density of solution `= 0.96 g mL^(-1)`) |
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Answer» Correct Answer - C::D `C_(2) H_(5) OH` is 20% by volume, i.e., `20 mL of C_(2) H_(5) OH` is dissolved in `100 mL` of solution Volume of `H_(2) O = 100 - 20 = 80 mL` Weight of `H_(2) O = 80 xx 1 = 80 g (d_(H_(2)O) = 1)` Weight of solution `= 100 xx 0.96 = 96 g` weight of `C_(2) H_(5) OH = 96 - 80 = 16 g` `(Mw of C_(2) H_(5) OH = 12 xx 2 + 5 + 16 + 1 = 46)` `:. M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)) = (16 xx 1000)/(46 xx 100) = 3.48` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (16 xx 1000)/(46 xx 80) = 4.35` |
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| 481. |
2.7 gm of Al is heated with 100 mL of `H_(2)SO_(4)` (29.4% w/w density 1 gm/mL) following reaction takes place: `Al+H_(2)SO_(4) rarr Al_(2)(SO_(4))_(3)+H_(2)` The volume of `H_(2)` gas evolved at 273 k and 1 atm:A. 3.36 LB. 2.24 LC. 4.48 LD. 11.2 L |
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Answer» Correct Answer - A |
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| 482. |
2.7 gm of Al is heated with 100 mL of `H_(2)SO_(4)` (29.4% w/w density 1 gm/mL) following reaction takes place: `Al+H_(2)SO_(4) rarr Al_(2)(SO_(4))_(3)+H_(2)` The hydrogen gas obtained in the reaction is mixed with `1.2 xx 10^(24)` molecules of `O_(2)` . The molecular weight of gaseous mixture is :(Assuming gases are not reacting `N_(0)=6 xx 10^(23))`A. 29.9B. 30.6C. 17D. None of these |
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Answer» Correct Answer - A |
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| 483. |
`5 mL` of a gaseous hydrocarbon was exposed to `30 mL` of `O_(2)`. The resultant gas, on cooling, is formed to measure `25 mL` of which `10 mL` is absrobed by `NaOH` and the remainder by pyrogallol. Determine the molecular formula of hydrocation. All measurements are made at constant pressure and temperature. |
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Answer» Let the formula of hydrocarbon `= C_(x) H_(y)` `:. C_(x) H_(y) + (x + (y)/(4)) O_(2) rarr xCO_(2) + (y)/(2) H_(2) O` `"Initial" 1 mL (x + (y)/(4)) mL 0 0` ltbgt `"Final" 5 mL 5 (x + (y)/(4)) mL 5x mLm -` `CO_(2) = 10 mL` `:. 5 x = 10 implies x = 2 mL` Volume of `O_(2)` left `= 15 mL` Volume of `O_(2)` used `= 30 - 15 = 15 mL`. `:. 5 (x + (y)/(4)) = 15 implies y = 4` Formula of hydrocarbon `= C_(2) H_(4)` |
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| 484. |
Which of the following reactions represents disproportionation ?A. `CrO_(5) rarr Cr^(3+)+O_(2)`B. `IO_(3)^(-)+I^(-)+H^(+)rarr I_(2)`C. `CrO_(2)Cl_(2)+NaOHrarr Na_(2)CrO_(4)+NaCl+H_(2)O`D. `Na_(2)S_(2)O_(3)+H_(2)SO_(2)rarr Na_(2)SO_(4)+SO_(2)+S_(8)+H_(2)O` |
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Answer» Correct Answer - 4 `Na_(2)S_(2)O_(3)+H_(2)SO_(4)rarr Na_(2)SO_(4)+SO_(2)+S_(8)+H_(2)O` Oxidation No. of `S` in `Na_(2)S_(2)O_(3)=+2` Oxidation No. of `S` in `H_(2)SO_(4)=+6` Oxidation No. of `S` in `SO_(4)=+4` Oxidation No. of `S` in `Na_(2)SO_(4)=+6` Oxidation No. of `S` in `S_(8)=0` So, Sulphur can oxidise from `+2` to `+4` as well as reduce to `0(S_(8))`. |
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| 485. |
The population of India based on 1981 census figure was 684 millions. Express the results in scientific notation and calculate the number of significant figures. |
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Answer» 1 Million `= 10^(6)` 684 Millions `= 684 xx 10^(6)` The value in terms of scientific notation may be expressed `6.84 xx 10^(8)` The number of significant figures = three. |
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| 486. |
Which of the statement are true?A. Law of constant compositon is true for all types of compounds.B. Molar volume of a gas at standard conditions is `22.4 L`.C. vapour density of a gas is twice of its molecula mass.D. Atomic masses of most elements are fractional. |
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Answer» Correct Answer - D a. Not applicable if the elements exists in different isotopes which may be involved in the formation of compound. b. At 1 atm, `25^(@)C` molar volume `= 22.7 L` c. `Mw = 2 xx VD`, Due ot existance of isotopes. |
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| 487. |
The largest number of molecules inA. `36 g` of waterB. `28 g` of carbon monoxideC. `46 g` of ethly alcoholD. `54 g` of nitrogen pentoxide |
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Answer» Correct Answer - A The number of molecules in `36 g` of water is `(36)/(18) N = 2 N` The number of molecules in `28 g` of `CO` is `(28)/(28) N = N` The number of molecules in `46 g` of `C_(2) H_(5) OH` is `(46)/(46) N = N` The number of molecules in `54 g` of `N_(2)O_(5)` is `(54)/(108) N = (1)/(2) N` |
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| 488. |
Which of the statement are true?A. The equivalent weight of `Ca_(3)(PO_(4))_(2)` is `Mw//6`.B. The equivalent weight of `Na_(3)PO_(4).12H_(2)O` is `Mw//3`.C. The equivalent weight of `K_(2) SO_(4)` is `Mw//2`.D. The equivalent weight of potas alum `K_(2)SO_(4)Al_(2)(SO_(4))_(3).24H_(2) O` is `Mw//8` |
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Answer» Correct Answer - A::B::C::D a. Valency factor `= 6 [overset(+ 2 xx 3)(Ca_(3)) overset(-3 xx 2)((PO_(4))_(2))]` b. Valency factor `= 3 [overset(+1 xx 3)(Na_(3)) overset(-3)((PO_(4)))]` c. Valency factor `= 2 [overset(+1 xx 2)(K_(2)) overset(-2)(SO_(4))]` d. Valency factor `= 8 [overset(+ 1 xx 2)(K_(2)) overset(-2)(SO_(4)) overset(+3 xx 2)(Al_(2)) overset(-2 xx 3)((SO_(4)))]` |
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| 489. |
A 50 gm oleum sample contains `(400/49)` gm of combined `SO_(3).` Find percent label of the oleum sample. |
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Answer» Correct Answer - 118 |
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| 490. |
An oleum sample has `SO_(3)` and `H_(2)SO_(4)` in 2:3 mass ratio. Select the correct statement(s).A. % labelling of sample is 109%B. % labelling of sample is 118%C. If 9 gm `H_(2)O` is added to200 gm of above sample, new labelling would be 104.5%D. If 9 gm `H_(2)O` is added to 200 gm of above sample, new labelling would be 104.3% |
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Answer» Correct Answer - A::D |
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| 491. |
The `Ew` of an element is `13`. It forms an acidic oxide which `KOH` forms a salt isomorphous with `K_(2)SO_(4)`. The `Aw` of element is a. `13`, b. `26`, c. `52`, d. `78`A. 13B. 26C. 52D. 78 |
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Answer» Correct Answer - D `K_(2) MSO_(4)` and `K_(2)SO_(4)` are isomorphous. Valency of `S` and `M` should be same `= 6` Atomic weight `= Ew xx` Valency `= 13.00 xx = 78` |
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| 492. |
The vapour density of a mixture containing equal number of moles of methane and ethane at `STP` isA. `11.5`B. `11.0`C. 23D. `12.0` |
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Answer» Correct Answer - A Molar mass of mixture will be `M_(mix)=(nxx"Molar mass of "CH_(4)+nxx"Molar mass of "C_(2)H_(6))/(n+n)` `M_(mix)=(n(16+30))/(2n)=23` Vapour density `rArr (M_(mix))/(2)=(23)/(2)=11.5` |
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| 493. |
For a reaction, `N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)` , identify dihtdrogen `(H_(2))` as a limiting reagent in the following reaction mixtures.A. `14g of N_(2) + 4g of H_(2)`B. `28g of N_(2) + 6g of H_(2)`C. `56g of N_(2) + 10g of H_(2)`D. `35g of N_(2) + 8g of H_(2)` |
| Answer» Correct Answer - C | |
| 494. |
A mixture of gases liberated upon decomposition of 33.1 gm of lead (II) nitrate is dissolved in 10 ml of water. What is the mass (in g) of 0.1 M KOH solution with density of 1.05 g/ml required to neutralize this acid. The reactions are: `2Pb(NO_(3))_(2)rarr2PbO+4NO_(2)+O_(2)` `KOH+HNO_(3)rarrKNO_(3)+H_(2)O` [Atomic mass of Pb=207] |
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Answer» Correct Answer - 2100 |
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| 495. |
The measured density at STP of He is 0.1762 g/L. What is the weightof one mole of He? |
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Answer» Correct Answer - 4 |
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| 496. |
Calculate the number of moles of ammonia required to produce 2.5 moles of `[Cu(NH_(3))_(4)]SO_(4).` |
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Answer» Correct Answer - 10 |
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| 497. |
Preparation of `Na_(2)SnO_(2)` involves the following set of reactions : `[Sn=119]` `(P) Sn+2HClrarrSnCl_(2)+H_(2)` `(Q)SnCl_(2)+2NaOHrarrSn(OH)_(2)+2NaCl` `(R)Sn(OH)_(2)+2NaOH rarrNa_(2)SnO_(2)+2H_(2)O` If `%` yield of reaction `(P),(Q),(R) 25 %,50 %,40 %` respectively. Calculate the mass of Sn (in kg required to produce 19.7 kg of `Na_(2)SnO_(2)`. |
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Answer» Correct Answer - 238 |
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| 498. |
A `6.90 M` solution of `KOH` contains 30% by weight of `KOH`. Calculate the density of the solution. |
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Answer» `M = (% "by weighy" xx 10 xx d)/(Mw_(2))` `6.9 = (30 xx 10 xx d)/(56)` `d = 1.288 g mL^(-1)` |
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| 499. |
In 2.6 gm of `FeSO_(4).6H_(2)O` (At wt of fe =56):A. No. of atoms of O are `100 xx N_(A)`B. Moles of H atoms are `120 xx N_(A)`C. Molecules of water are `60 xx N_(A)`D. Moles of `e^(-)` present in `SO_(4)^(2-)` are 500 |
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Answer» Correct Answer - A::C::D |
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| 500. |
Calculate the volume of `O_(2)` and volume of air needed for combustion of `1 kg` carbon at `STP`. |
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Answer» `underset(12 g)(C) + underset(32 g)(O_(2)) rarr CO_(2)` `12 g` of `C implies g` of `O_(2) = 1 mol` of `O_(2) = 22.4 L` at `STP` `1000 g` of `C = (22.4 xx 1000)/(12) = 1866.67 L "of" O_(2)` Air contains `= (20 % "of" O_(2) + 80% "of N_(2))` Volume of air = 5 times volume of `O_(2) = 5 xx 1866.67` `= 9333.33 L` |
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