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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.
| 51. |
A gases mixture contains oxygen and nitrogen in the ratio `1 : 4` by weight. Therefore, the ratio of the number of molecules is:A. `1 : 4`B. `1 : 8`C. `7 : 32`D. `3 : 16` |
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Answer» Correct Answer - C Mol. Ratio `O_(2):N_(2)` `(1)/(32) : (4)/(28)` Molecules `(N_(A))/(32) : (4N_(A))/(28) implies 7:32` |
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| 52. |
A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is labelled 86 percent by weight. What is the molarity of the solution ? What volime of the acid is required to make it 1 litre of 0.2 M `H_(2)SO_(4)` ? |
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Answer» Correct Answer - (i) 15.68 M (ii) 12.75 mL Step I. Molarity of solution Mass of `H_(2)SO_(4)=86 g` , Mass of the solution = 100 g Volume of 100 g of the solution `= ("Mass")/("Density") = (100)/(1.787) = 55.96 mL = 0.05596 L` Molarity of solution (M) `= (("Mass of "H_(2)SO_(4))/("Molecular mass of "H_(2)SO_(4)))/("Volume of solution in mL"/(1000))=((86g))/(("98 g mol"^(-1))xx("0.05596 L"))` `=15.68 "mol L"^(-1)=15.68 M` Step II. Volume of solution required `overset(("conc."))(M_(1)V_(1))-=overset(("dilute"))(N_(2)V_(2))` `15.68xxV_(1)=0.2xx1000,V_(1)=(0.2xx1000)/(15.68)=12.75mL`. |
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| 53. |
A sample of `H_(2) SO_(4)` (density `1.787 g mL^(-1)`) si labelled as 86% by weight. What is the molarity of acid? What volume of acid has to be used to make `1 L` of `0.2 M H_(2) SO_(4)` ? |
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Answer» `H_(2) SO_(4)` is 86% by weight `:.` weight fo `H_(2) SO_(4) = 86 g` Weight of solution `= 100 g` `:.` Volume of solution `= (100)/(1.787) mL = (100)/(1.787 xx 1000) L` `M_(H_(2)SO_(4)) = (86)/(98 xx (100)/(1.787 xx 1000)) = 15.68` Let `V mL` of this `H_(2) SO_(4)` are used to prepare `1 L` of `0.2 M` `H_(2) SO_(4)` `:.` mmoles of concentrated `H_(2) SO_(4)` = mmoles of dilute `H_(2) SO_(4)` `:. V xx 15.68 = 1000 xx 0.2` `:. V = 12.75 mL` |
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| 54. |
A bottle of commercial sulphuric acid `(d = 1.787 g mL^(-1))` is 86% by weight. a. Whatis molarity of the acid? b. What volume of the acid has to be used to make `1 L` of `0.2 M H_(2) SO_(4)`? c. What is the molality of the acid? |
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Answer» a. `M = (% "by weight" xx 10 xx d)/(Mw_(2)) = (86 xx 10 xx 1.787)/(98) = 15.68` b. `M_(1) V_(1) = M_(2) V_(2)` `15.68 xx V_(1) = 0.2 xx 1` `V_(1) = 0.01274 L = 12.74 mL` c. `W_(1) =` Weight of `H_(2) O` = Weight of solution - Weight of solute `= 100 - 86 = 14 g` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))` `= (86 xx 1000)/(98 xx 14) = 62.68` |
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| 55. |
What is the percentage composition of each element is zinc-phosphate `Zn_(3) (PO_(4))_(2)`? `(Zn = 65.5, P = 31, O = 16)` |
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Answer» `Mw` of `Zn_(3) (PO_(4))_(2) = 65.5 xx 3 + 2[31 + 16 xx 4]` `= 196.5 + 2 xx 95 = 386.5` `% of Zn = (196.5 xx 100)/(386.5) = 50.84%` `% of P = (62 xx 100)/(386.5) = 16.04%` `% of O = (128 xx 100)/(386.5) = 32.12%` |
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| 56. |
A solid mixture `5 g` consists of lead nitrate and sodium nitrate was heated below `600^(@)C` until weight of residue was constant. If the loss in weight is `28%` find the amount of lead nitrate and sodium nitrate in mixture. |
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Answer» Mass of solid mixture = 5.0 g Let the mass of lead nitrate = x g `:.` Mass of sodium nitrate `= (5-x)g` Loss in mass of mixture `= (5xx28)/(100)=1.4g` `:.` Mass of the residue `= 5 - 4 = 3.6 g` Calculation of the mass of the residue from lead nitrate. `underset(underset("= 663 g")(2(207+28+96)))(2Pb(NO_(3))_(2))rarrunderset(underset(=446g)(2(207+16)))(2PbO)+4NO_(2)+O_(2)` 662 g of `Pb(NO_(3))_(2)` form residue (PbO) = 446 g `:. x` g of `Pb(NO_(3))_(2)` form residue `(PbO)=(446)/(662) xx xg` Step II. Calculation of the mass of the residue from sodium nitrate `underset(underset(=170g)(2(23+14+48)))(2NaNO_(3))rarrunderset(underset(=138g)(2(23+14+32)))(2NaNO_(2)+O_(2))` 170g of `NaNO_(3)` form residue `(NaNO_(2))=138g` `:. (5-x)g` of `NaNO_(3)` form residue `(NaNO_(2))=(138)/(170) xx (5-x)g` Step III. Calculation of the amount of lead nitrate and sodium nitrate Total mass of PbO and `NaNO_(3)=[(446)/(662)x+(138)/(170)(5-x)]g` Total mass of the residue actually left = 3.6 g `:. (446)/(662)x+(138)/(170)(5-x)=3.6` `0.674x+0.812(5-x)=3.6` or `0.674x+4.06 -0.812x = 3.6` or `0.674x-0.812x = 3.6 - 4.06` `-0.13x = - 0.46` `:. x = (0.46)/(0.138)=3.38g` Thus, Mass of `Pb(NO_(3))_(2)=3.38g` Mass of `NaNO_(3)=5 - 3.38 = 1.62g`. |
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| 57. |
Zinc and hydrochloric acid react according to the reaction: `Zn_((s))+2HCl_((aq.))rarr ZnCl_(2(aq.))+H_(2(g))` If `0.30` mole of `Zn` are added to hydrochloric acid containing `0.52` mole `HCl`, how many moles of `H_(2)` are produced? |
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Answer» The balanced chemical equation for the reaction is : `underset("1 mol")(Zn(s))+underset("2 mol")(2HCl(aq))rarrZnCl_(2)(aq)+underset("1 mol")(H_(2)(g))` Step I. Determine of limiting reactant 1 mole of Zn react with HCl = 2 mol 0.30 mole of Zn react with HCl `= 2 xx 0.3 = 0.6 mol` But HCl actually present in solution `= 0.52` mol This means that HCl is present in lesser amount than what is actually required. `:.` It is the limiting reactant Step II. Determination of moles hydrogen produced 2.0 mole of HCl on reacting with Zn produce `H_(2)(g)=1` mol 0.52 mole of HCl on reacting with Zn produce `H_(2)(g)=(1xx 0.52)/(2)=0.26` mol. |
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| 58. |
Statement-I : Pure water obtained from different sources such as, river, well, spring, sea etc. always contains hydrogen and oxygen combined in the ratio `1 : 8` by mass. Because Statement-II : `A` chemical compound always contains elements combined together in same proportion by mass. it wa discovered by French chemist, Joseph Proust `(1799)`.A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true , Statement-II is `NOT` a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - A | |
| 59. |
Mole fraction of `A` in `H_(2)O` is `0.2` The molality of `A` in `H_(2)O` is :A. `13.8`B. `15.5`C. `14.5`D. `16.8` |
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Answer» Correct Answer - A `x_(Lambda)=0.2` `x_(H_(2))O = 1-0.2=0.8` `"wt of" H_(2)O = 0.8xx18 = 14.4 g` Molality `=("moles of solute")/("wt. of solvent" (H_(2)O)"in kg")` `= (2xx1000)/(14.4)=13.8` |
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| 60. |
Assertion: The weight percentage of a compound A in a solution is given by `% of A=("Mass A")/("Total mass of solution")xx100` Reason: The mole fraction of a component A is given by, Mole fraction of A `=("No. of moles of A")/("Total no. of moles of all components")`A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true , Statement-II is `NOT` a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - B | |
| 61. |
How many blood cells of 5 ml each having `[K^(+)]=0.1 M` should burst into 25 ml of blood plasma `[K^(+)]=0.02 M` so as to give final `[K^(+)]=0.06 M` ? |
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Answer» Correct Answer - 5 |
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| 62. |
What is molarity of the resulting solution obtained mixing `2.5 L` of 0.5 M urea solution and 500 mL of 2M urea solution ? |
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Answer» Molarity of solution mixing : `M_(1)V_(1)+M_(2)V_(2)=M_(3)V_(3)` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` According to available data : `M_(1)-0.5 M,V_(1)=2.5L,M_(2)=2M,V_(2)=500 mL=0.5 L` `:. M_(3)=((0.5 Mxx2.5L)+(2Mxx0.5L))/((2.5L+0.5L))=((2.25ML))/((3.0L))=0.75M`. |
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| 63. |
What is the volume of ethyl alcohol (density 1.15 g/cc) that has to be added to prepare 100 cc of 0.5 M ethyl alcohol solution in water ? |
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Answer» Molarity of solution (M) = `(("Mass of ethyl alcohol")/("Molar mass"))/("Volume of solution in dm"^(3))` `(0.5 "mol dm"^(-3))=("Mass of ethyl alcohol")/((46" g mol"^(-1))xx(0.1 "dm"^(3)))` Mass of ethyl alcohol `(C_(2)H_(5)OH)=(0.5 "mol dm"^(-3))xx(46 "g mol"^(-1))xx(0.1 "dm"^(3))` Volume of ethyl alcohol solution `= ("Mass")/("Density")=((2.3 g))/((1.15 "g/cc"))=2.0` cc. |
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| 64. |
Assertion: The weight percentage of a compound A in a solution is given by `% of A=("Mass A")/("Total mass of solution")xx100` Reason: The mole fraction of a component A is given by, Mole fraction of A `=("No. of moles of A")/("Total no. of moles of all components")`A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is Ture, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - B |
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| 65. |
Haemoglobin`C_(2952)H_(4664)N_(812)O_(832)S_(8)Fe_(4)`, molar mass = 65248 g/mol is the oxygen carrier in blood. An average adult has about 5.0L of blood. Every milliliter of blood has approximately `6.0 xx 10^(9)` erythrocytes, or red blood cells and every red blood cell has about `3 xx 10^(8)` haemoglobin molecules. The mass of haemoglobin molecules in an average adult is : `(N_(A) = 6 xx 10^(23))`A. 978.72gmB. 652.48gmC. 434.99 gmD. 0.015gm |
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Answer» Correct Answer - A |
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| 66. |
100 mL of 0.06 M`Ca(NO_(3))_(2)` is added to 50 mL of 0.06 M `Na_(2)C_(2)O_(4)` . After the reaction is complete.A. 0.003 moles of calcium oxalate will get precipatatedB. 0.003 M of `Ca^(2+)` will remain in excessC. `Na_(2)C_(2)O_(4)` is limited reagentD. `Ca(NO_(3))_(2)` is excess reagent |
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Answer» Correct Answer - A::C::D |
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| 67. |
The molarity of the solution containing 2.8% (mass/volume) solution of KOH is: (Given atomic mass of K=39) is:A. `0.1M`B. `0.5M`C. `0.2M`D. `1M` |
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Answer» Correct Answer - B `2.8%` by mass volume solution of KOH i.e., `2.8 g KOH` in `100ml` solution molarity `=(2.8)/(56xx.1)=.5M` |
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| 68. |
A compound contains 5 gm sulphur and 5 gm oxygen atom. The empirical formula of compound is:A. SOB. `SO_(2)`C. `S_(2)O`D. `SO_(3)` |
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Answer» Correct Answer - B |
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| 69. |
Density of a 3 molar aqueous solution of `Na_(2)S_(2)O_(3)` is `1.482gm//ml`. Calculate mole fracrtion of `Na_(2)S_(2)O_(3)` in solution.A. `0.054`B. `0.06`C. `0.03`D. `0.072` |
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Answer» Correct Answer - A |
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| 70. |
The molarity of the solution containing `2.8%` (mass/volume) solution of KOH is : (Given atomic mass of K = 39)A. `0.1` MB. `0.5` MC. `0.2` MD. 1 M |
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Answer» Correct Answer - B |
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| 71. |
Statement-1 : When 40 gm of NaOH is mixed with 49gm `H_2SO_4` and mixed with water then 89gm of `NaSO_4` is obtained assuming `100%` yield. Statement-2 : For producing maximum amount of `Na_2SO_4` with `100%` yield NaOH and `H_2SO_4` should be present in a molar ratio of `2:1`.A. Statement-1 : is True, Statement-2 : is True, Statement-2 : is a correct explanation for Statement-1 :B. Statement-1 : is True, Statement-2 : is True, Statement-2 : is NOT a correct explanation for Statement-1 .C. Statement-1 : is True, Statement-2 : is False.D. Statement-1 : is False, Statement-2 : is True. |
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Answer» Correct Answer - D |
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| 72. |
How many neutrons are in 0.025 molof the isotope `._24^54Cr` ?A. `1.5 xx 10^(22)`B. `3.6 xx 10^(23)`C. `4.5 xx 10^(23)`D. `8.1 xx 10^(23)` |
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Answer» Correct Answer - C |
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| 73. |
The number of neutrons in 0.45g water, assuming that all the hydrogen atoms are `H^(1)` atoms and all the oxygen atoms are `O^(16)` atoms, is:A. 8B. 0.2C. `1.2 xx 10^(23)`D. `4.8 xx 10^(24)` |
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Answer» Correct Answer - C |
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| 74. |
What is the molarity of `H_(2)SO_(4)` solution that has a density of `1.84g//cc` and contains `98%` by mass of `H_(2)SO_(4)`?A. `4.18` MB. `8.14` MC. `18.4` MD. 18 M |
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Answer» Correct Answer - C |
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| 75. |
`H_(3)PO_(4) (98 g mol^(-1))` is 98% by mass of solution. If the density is 1.8 g/ml, the molarity is:A. 18 MB. 36 MC. 54 MD. `018` M |
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Answer» Correct Answer - A |
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| 76. |
The numberof nitrogen atoms in 3.68g of `K_(4)[Fe(CN)_(6)]` is: [`N_(0)` = Avogadro number]A. 0.06B. `0.01 N_(0)`C. `0.06 N_(0)`D. none of these |
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Answer» Correct Answer - C |
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| 77. |
A typical polyethylene bag froma grocery store weighs `12.4g`. How many molecules of ethylene, `C_(2)H_(4)`, must be polymerized to make such a bag?A. `1.36 xx 10^(24)`B. `6.02 xx 10^(23)`C. `5.33 xx 10^(23)`D. `2.67 xx 10^(23)` |
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Answer» Correct Answer - D |
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| 78. |
What is the maximum mass of (in grams) of NO that could be obtained from 15.5g of `N_(2)O` and 4.68g of `N_(2)H_(4)` when they react? The balanced chemical equation is: `2N_(2)O_(4) + N_(2)H_(4) rArr 6NO + 2H_(2)O` `{:("Molar mass" , (gmol^(-1))), (N_(2)O_(4), 92.0), (N_(2)H_(4), 32.0):}`A. 4.38B. 5.04C. 15.2D. 26.2 |
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Answer» Correct Answer - C |
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| 79. |
Mr. Gupta has lost the secret code of his bag which consists of lots of chocolates.From the information given below help Mr. Gupta to recall his code. The code consists of five digits a b c d e : (a) = represents moles of hydrogen gas formed by converting all the hydrogen in 6 moles of `NH_(3)` (b) `= ("density of SO"_(2) "gas at same T and P")/("density of O"_(2)"gas at same T and P")` (c) `= %` moles of `NH_(3)` in a mixture of `NH_(3)` and `H_(2)S` having an average molecular weight of 33.15 (d,e) = represents `%` yield of reaction if 16.8 L of `O_(2)` is produced at 1 atm and 273 K from `122.5 gm` of `KClO_(3)`. |
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Answer» Correct Answer - 92550 |
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| 80. |
A mixture weight of glucose-1-phosphate is 260 and its density is `1.5g//ml`. What is the average volume occupied by 1 molecular of this compound?A. 24B. 20C. 26D. 40 |
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Answer» Correct Answer - A |
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| 81. |
One mole of a mixture of `N_(2),NO_(2)` and `N_(2)O_(4)`, has a mean molar mass of `55.4`. On heating to a temperature at which `N_(2)O_(4)` may be dissociated `:N_(2)O_(4)rarr2NO_(2)`, the mean molar mass tends to the lower value of `39.6`. What is the mole ratio of `N_(2) : NO_(2) : N_(2)O_(4)` in the original mixture? |
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Answer» Correct Answer - `0.5 : 0.1 : 0.4` Mixture `(N_(2),NO_(2),N_(2),O_(4))` has mean molar mass `=55.4` `x" "y" "z` Given: `{:(N_(2)O_(4)rarr2NO_(2)),(z" "2z):}` `therefore 55.4=(28x+46(y+2z))/(x+y+z)` `{"mean molar mass" =(wtxx"mole")/("Total mole")}` Given : `x+y+z =1` (mole) so `55.4 = 28x + 46 (y+2z)" "...(1)` `therefore 39.3 = (28x + 46(y+2z))/(x+y+2z)` `therefore 39.6(x+y+2z)=28x+46(y+2z)` From `eq (1) & x+y+z=1` or `1+z = (59.4)/(39.6)` or `z=0.4` from `eq. (1)` `55.4 = 28 x+46(y+2z)` `z=0.4` ` 55.4 = 28x + 46y +36.8` `28x+46y=18.6 " "(2)` ` because x+y+z=1` `x+y+0.4=1(beacuse z =0.4)` `x+y=0.6 " "...(3)` `eq (2) xx1....eq. (3) xx28` `{:(,28x+46y=18.6),(,28x+28y=16.8),(-, " - -"),(,bar(" 18y=1.8")):}` `y=0.1` `because x+y+z=1 x=0.5` |
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| 82. |
Molality of pure liquid ethanol `(C_(2)H_(5)OH)` if its density `d=1.2g//ml` is :A. `0.83` mB. 50 mC. `0.78` mD. `21.74` m |
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Answer» Correct Answer - D |
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| 83. |
What is the specific gravity of a liquid if 260 mL of the liquid has the same mass as 390 mL of water?A. `0.66`B. `0.5`C. `1.5`D. `1.8` |
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Answer» Correct Answer - C |
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| 84. |
One mole of a mixture of `N_(2),NO_(2)` and `N_(2)O_(4)`, has a mean molar mass of `55.4`. On heating to a temperature at which `N_(2)O_(4)` may be dissociated `:N_(2)O_(4)rarr2NO_(2)`, the mean molar mass tends to the lower value of `39.6`. What is the mole ratio of `N_(2) : NO_(2) : N_(2)O_(4)` in the original mixture?A. `0.5 :0.1:0.4`B. `0.6:0.1:0.3`C. `0.5 : 0.2 :0.3`D. `0.6 : 0.2 : 0.2` |
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Answer» Correct Answer - A |
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| 85. |
45 g of gaseous mixture of `NO_(2) N_(2)O` gas occupy 22.7 litre at STP. Find the mole `%` of `NO_(2)` in the gaseous mixture. |
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Answer» Correct Answer - 50 |
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| 86. |
8 mole of a mixture of `N_(2), NO_(2)` and `N_(2)O_(4)` has a mean molecular mass of `(378)/(8)` . On heating to a temperature at which `N_(2)O_(4)` dissociated completely `(N_(2)O_(4)rarr 2NO_(2))` , the mean molecular mass become `(378)/(9)`. The ratio of number of moles of `N_(2): NO_(2):N_(2)O_(4)` in original mixture is :A. `3:4:1`B. `2:5:1`C. `3:5:2`D. `4:2:3` |
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Answer» Correct Answer - B |
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| 87. |
8 mole of a mixture of `N_(2), NO_(2)` and `N_(2)O_(4)` has a mean molecular mass of `(378)/(8)` . On heating to a temperature at which `N_(2)O_(4)` dissociated completely `(N_(2)O_(4)rarr 2NO_(2))` , the mean molecular mass become `(378)/(9)`. The ratio of number of moles of `N_(2)` and `NO_(2)` after heating is :A. `1:1`B. `3:9`C. `2:7`D. `1:2` |
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Answer» Correct Answer - C |
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| 88. |
Based on following reaction given, minimum possible value of `x` for 1 mol `AF_(6)` will be : `AF_(6)+H_(2)OrarrAO_(x)+HF`. |
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Answer» Correct Answer - 3 |
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| 89. |
`N_(2)O_(4)` dissociates into `NO_(2)`. If `%` dissociation of `N_(2)O_(4)` is `33.33 %`, calculate average molecular weight of gaseous mixture formed. |
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Answer» Correct Answer - 69 |
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| 90. |
`0.41g` of the silver salt of a dibasic organic acid left a residue to `0.216` of silver on ignition. Calculate the molecular mass of the acid . |
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Answer» Mass of the silver salt taken `(W) = 0.41g` Mass of Ag fromed `=0.216g` `H_(2)Xrarrunderset(w=0.41g)(Ag_(2)X)rarrunderset(x=0.216g)(2Ag)` Now malar mass acid `=n((108W)/(x)-107)gmol^(-1)=2((108xx0.41)/(0.216)-107)gmol^(-1)=196mol^(-1)` Moleclar mass `= 196 g//mol` . |
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| 91. |
Sodium is an essential constituent of our body. Calculate the percentage of sodium in the breakfast cereal which has been advertised to contain 40 mg of sodium per 50 g of the cereal. |
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Answer» Percentage of sodium `= ("Mass of sodium grams")/("Mass of cereal in grams")xx100` `=((40xx10^(-3)g))/((50.0g))xx100=0.08%`. |
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| 92. |
One litre of `N//2 HCl` solution was heated in a beaker. When the volume was reduced to `600 mL`, `9.125 g` of `HCl` was lost out the new normality of solution is a. `~~ 0.4` b. `~~ 0.8` c. `~~ 0.4` d. `~~ 0.2` |
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Answer» `underset({:(0.5 N xx 1 L),(= 0.5 Eq),(= 0.5 xx 363.5 g),(= 18.25 g):})(HCl) overset(Delta)rarr underset(600 mL)(HCl)` Mass of `HCl` left after heating `= 18.25 - 9.125` `= 9.125 g` `N_("new")` of `HCl = (9.125 xx 1000)/(36.5 xx 600) = 0.416 N` |
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| 93. |
`10 g` mixture of `NaHCO_(3)` and `Na_(2)CO_(3)` has `1.68 g NaHCO_(3)`. It is heated at `400 K`. Weight of the residue will beA. `9.38 g`B. `8.32 g`C. `10.0 g`D. `1.68 g` |
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Answer» Correct Answer - A `Na_(2) CO_(3)` is not affected by heating `2NaHCO_(3) rarr Na_(2) CO_(3) + H_(2) O + CO_(2)` Residue |
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| 94. |
`1 g` of calcium was burnt in excess of `O_(2)` and the oxide was dissolved in water to make up `1 L` solution. Calculate the normality of alkaline soluiton. |
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Answer» `Ca + (1)/(2) O_(2) rarr CaO` `40 g of Ca = 56 g of CaO` `1 g of Ca = (56)/(40) = 1.4 g of CaO` `N = (W_(2) xx 1000)/(Ew xx V_(sol) ("in" mL)) = (1.4 xx 1000)/((56)/(2) xx 1000) = 0.05 N` |
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| 95. |
24 gm pure sample of magnesium is burnt in air to form magnesium oxide and magnesium nitride. When products are treated with excess of `H_(2)O` , 3.4 gm of gaseous `NH_(3)` is generated according to given reactions. `Mg+O_(2) rarr MgO` `Mg+N_(2)rarr Mg_(3)N_(2)` `Mg_(3)N_(2)+ 6H_(2)O rarr 2Mg(OH)_(2)+2NH_(3)` Calculate the amount of `Mg(OH)_(2)` (in gm) produced in above reaction.A. 11.6B. 17.4C. 23.2D. 15.8 |
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Answer» Correct Answer - B |
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| 96. |
24 gm pure sample of magnesium is burnt in air to form magnesium oxide and magnesium nitride. When products are treated with excess of `H_(2)O` , 3.4 gm of gaseous `NH_(3)` is generated according to given reactions. `Mg+O_(2) rarr MgO` `Mg+N_(2)rarr Mg_(3)N_(2)` `Mg_(3)N_(2)+ 6H_(2)O rarr 2Mg(OH)_(2)+2NH_(3)` Calculate the amount of magnesium oxide (in gm) in products.A. 28B. 20C. 16.8D. 32 |
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Answer» Correct Answer - A |
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| 97. |
Pure carbon was burnt in excess of oxygen . The gaseous products are `CO_(2)=60` mole% CO=15 mole % , `O_(2)` =25 mole% In the above gaseous mixture if `SO_(2)` gas is introduced then which of the following will be correct?A. Molar mass of mixture will increaseB. Gaseous mixture will become heavierC. (a) and (b) both are correct.D. Molar mass remains same for homogenous system |
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Answer» Correct Answer - C |
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| 98. |
100 g a calcium was burnt in excess of `O_(2)` and the oxide obtained was dissolved in water to make 1 litre solution. Calculate the molarity of `OH^(-)` ion of the alkaline solution. |
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Answer» Correct Answer - 5 |
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| 99. |
`C_(6)H_(5)OH(g)+O_(2)rarrCO_(2)(g) +H_(2)O(l)` Magnitude of volume change if `30ml` of `C_(6)H_(5)OH (g)` is burnt with excess amount of oxgen, isA. 30 mlB. 60 mlC. 20 mlD. 10 ml |
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Answer» Correct Answer - B |
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| 100. |
`3.2 g` sulphur combines with `3.2 g` of oxygen, to from a compound in one set of conditions. In another set of conditions `0.8 g` of sulphur combines with `1.2 g` of oxygen to form another compound. State the law illustrated by these chemical combinations. |
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Answer» Correct Answer - A First case `3.2 g` of `S` combines with `3.2 g of O_(2)` `1 g` of `S` combines with `= 1 g of O_(2)` Second case `0.8 g` of `` combines with `1.2 g of O_(2)`. `1 g` of `S` combines with `= (1.2)/(0.8) = 1.5 g of O_(2)` Thus, the ratio of the `O_(2)` in both cases which combines with a fixed mass `(1 g)` of `S = 1 : 15` or `2 : 3`, which is a simple whole number ratio, and hence the law of multiple porportion is verified. |
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