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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.
| 51. |
What are polyatomic ions? Give examples. |
| Answer» A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example,Nitrate `(NO_(3))`, hydroxide ion `(OH^(−))`. | |
| 52. |
How many moles are represented by 100 g of glucose. `C_(6)H_(12)O_(6)` ? `(C = 12 u , H = 1 u , O = 16 u)` |
| Answer» Correct Answer - `0.55` mol | |
| 53. |
The molecular formula of glucose is `C_(6)H_(12)O_(6)`. Calculate its molecule mass. (Atomic masses : `C = 12 u , H = 1 u , O = 16 u`) |
| Answer» Correct Answer - 180 u | |
| 54. |
Write the formulae for the following and calculate the molecular mass for each one of them. a) Caustic potash b) Baking powder c) lime stone d) caustic soda e) Ethanol f) Common salt |
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Answer» `(a)KOH=39+16+1=54u` `(b)NaHCO_(3)=23+1+12+3xx16=84u` `(c)CaCO_(3)=12+12xx3xx16=100u` `(d)NaOH=23+16+1=40u` `(e)C_(2)H_(5)OH=2xx12+6xx1+16=46u` (f)`NaCl=23+35.5=58.5u` |
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| 55. |
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? |
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Answer» The chemical reaction leading to products is `"sodium hydrogen carbonate"+"ethanoic acid" to underset("residue left")("sodium ethanoic solution")+underset("released")("carbon dioxide")` `"Mass of reactants"=(6.3+15.0)=21.3g` `"Mass of products"="Mass of residue"+"Mass of carbon dioxide released"` =18.0+x g According to law of conservation of mass. Mass of reactants=Mass of products 21.3g=(18.0+x)g or x=21.3-18.0=3.3g Mass of carbon dioxide released=3.3g |
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| 56. |
Which of the following represents a correct chemical formula? Name it.A. `CaCl`B. `BiPO_(4)`C. `NaSO_(4)`D. NaS |
| Answer» Only the chemical formula `BiPO_(4)` is correct (b). The correct formulae for the other compounds are: (a) `CaCl_(2)` (c )`N_(2)SO_(4)` (d) `Na_(2)S` | |
| 57. |
Why is it not possible to see an atom with naked eyes? |
| Answer» It is not possible to see an atom with naked eye because of its extremely small size. For example, the radius of an atom of hydrogen is of the order of `10^(-10)m`. Actually an atom is regarded as a microscopic particle. These microscopic particles cannot be seen with naked eye. | |
| 58. |
Which of the following represents the correct chemical formula? `(i) NaSO_(4), (ii) CaPO_(4) (iii) ZnS (iv) AlSO_(4)` |
| Answer» The formula (c) represents the correct formula. Both the ions are divalent i.e. `Zn^(2+) and S^(2-)`. The name of the Compound is zinc sulphide. | |
| 59. |
Calculate the formula unit mass of `CaCl_(2)`. |
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Answer» Formula unit mass of `CaCl_(2)`=Calcium chloride `=(1xx"Atomic mass of Ca")+(2xx"Atomic mass of Cl")` `=(1xx40u)+(2xx35.5u)=111u` |
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| 60. |
Calculate the number of molecular of sulphur `(S_(8))` present in 128g of sulphur. |
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Answer» `256"g of sulphur "(S_(8))"represent"=1"mol"` `128"g of sulphur "(S_(8))"represent"=((1"mol")xx(128g))/((256g))=0.5"mol"` `"1 mole of "S_(8)"has molecules"=6.022xx10^(23)` 0.5 moles of `S_(8)` has molecules `=((0.5"mol"))/((1.0"mol"))xx6.022xx10^(23)=3.011xx10^(23)` |
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| 61. |
Calculate the number of molecules of `SO_(2)` present in 44g of it. |
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Answer» Gram molecular mass of `SO_(2)=32xx2xx16=64g` `64"g of "SO_(2)"have molecules"=N_(A)=6.022xx10^(23)` `44"g of "SO_(2)"have molecules"=((6.022xx10^(23)))/((64g))xx(44g3.76xx10^(23))` |
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| 62. |
Calculate formula unit mass of `Na_(2)CO_(3).10H_(2)O` |
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Answer» Formula unit mass of `Na_(2)CO_(3).10H_(2)O` `2xx"Atomic mass of Na"+"Atomic mass of C"+3xx "Atomic mass of O"+10xx"Molecular mass of "H_(2)O` `=2xx23+12+3xx16+10xx19=286u` |
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| 63. |
Name the following compounds. Also write the symbols/formulae of the ions present in them : (a) `CuSO_(4)` , (b) ` (NH_(4))_(2)SO_(4)` , (c ) `Na_(2)O` , (d) `Na_(2)CO_(3)` , (e ) `CaCl_(2)` |
| Answer» Correct Answer - (a) Copper sulphate ; `Cu^(2+)` and `SO_(4^(2-))` , (b) Ammonium sulphate ; `NH_(4^(+))` and `SO_(4^(2-))` (c ) Sodium oxide ; `Na^(+)` and `O^(2-)` (d) Sodium carbonate ; `Na^(+)` and `CO_(2^(3-))` (e ) Calcium chloride ; `Ca^(2+)` and `Cl^(-)` | |
| 64. |
Four samples of water `[H_(2)O]` are collected from different sources, Each sample on analysis was found to containsame percentage of oxygen. Which law of chemical combination is demonstrated by the above observation? |
| Answer» The law of constant combination is demonstrated by this observation. | |
| 65. |
If the valency of hydrogen is 1 and that of nitrogen is 3, work out the formula for ammonia. |
| Answer» Correct Answer - `NH_(3)` | |
| 66. |
An liquid compound X of molecule mass 18 u can be obtained from a number of natural sources. All the animals and plants need liquid X for their survival. When an electric current is passed through 200 grams of pure liquid X under suitable conditions, then `178` grams of gas Y and 22 grams of gas Z are produced. Gas Y is produced at the positive electrode whereas gas Z is obtained at the negative electrode. Moreover, gas Y supports combustion whereas gas Z burns itself causing explosions. (a) Name (i) liquid X (ii) gas, Y and (iii) gas Z. (b) What is the ratio of the mass of element Z to mass of element Y in the liquid X? (c ) Which law of chemical combination is illustrated by this example ? (d) Name two source of liquid X. (e ) State an important use of Y in our life. |
| Answer» Correct Answer - (a) (i) Water , (ii) Oxygen (iii) Hydrogen , (b) `1 : 8` , (c ) Law of constant proportions (d) Rivers and Wells , (e ) Gas Y (oxygen) is necessary for brething | |
| 67. |
One of the forms of a naturally occurring solid compound P is usually used for making the floors of houses. On adding a few drops of dilute hydrochloric acid to P, brisk effervescence on adding dilute HCl to P. Gas Q is said t cause globel warning wheres solid R is used for white-washing. (a) What is (i) solid P (ii) gas Q, and (iii) solid R. (b) What is the total mass of Q and R obtained from 50 g of P ? (c) How does the total mass of Q and R formed compare with the mass of P taken ? (d) What conclusion do you get from the comparison of masses of products and reactant ? (e ) Which law of chemical combination is illustrated by the example given in this problem ? |
| Answer» Correct Answer - (a) (i) Calcium carbonate `(CaCO_(3))` in the form of marble (ii) Carbon dioxide `(CO_(2))` (iii) Calcium oxide `(CaO)` (b) `50 g` (c ) Total mass of Q and R `(50 g)` is euqal to the mass of P taken `(50 g)` (d) The mass of products is equal to the mass of reactant (e ) Law of conservation of mass | |
| 68. |
Compute the number of ions present in 5.85g of sodium chloride. |
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Answer» Gram formula mass of NaCl=23+35.5=58.5g 58.5g of NaCl have ions=`2xxN_(A)` 5.85g of NaCl have ions=`=(2xxN_(A)xx(5.85g))/((58.5g))` `=(2xx6.022xx10^(23))/(10)=1.2042xx10^(23)"ions"` |
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| 69. |
What do we call those particles which have : (a) more electrons than the normal atoms ? (b) less electrons than the normal atoms ? |
| Answer» Correct Answer - (a) Anions , (b) Cations | |
| 70. |
What do we call those particles which are formed : (a) by the gain of electrons by atoms ? (b) by the loss of electrons by atoms ? |
| Answer» Correct Answer - (a) Anions , (b) Cations | |
| 71. |
An element has valency 3. Write the formula of its oxide. |
| Answer» The valency of oxygen is 2 while that of element E is 3. The formula of the oxide of the element is: `E_(2)O_(3)`. | |
| 72. |
An element X has a valency of 2. Write the simplest formula for : (a) bromide of the element (b) oxide the element |
| Answer» Correct Answer - (a) `XBr_(2)` , (b) `XO` | |
| 73. |
What is the mass of `0.5` mole of wate `(H_(2)O)`. (Atomic masses : `H = 1 u, O = 16 u`) |
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Answer» In order to solve this problem, we should know the mass of 1 mole of water. This can be obtained by using the given values of the atomic masses of hydrogen and oxygen as follows : `1` mole of water ` (H_(2)O)` = Molecular mass of `H_(2)O` in grams = Mass of 2H atoms + Mass of H atom `= 2 xx 1 + 16` `= 2 + 16` `= 18` grams Thus, the mass of 1 mole of water is 18 grams. Now, Mass of 1 mole of water `= 18` g So, Mass of `0.5` mole of water `= 18 xx 0.5 g` `= 9 g` Thus, the mass of 0.5 mole of water `(H_(2)O)` is `9` grams. |
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| 74. |
How many gram molecules of `H_(2)SO_(4)`, are present in 4.9 g of the acid? (b) How many atoms of hydrogen and oxygen are present in 0-15 mole of water `(H_(2)O)`? |
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Answer» (a) Gram molecular mass of `H_(2)SO_(4)=2xx1+32+4xx16=98g` `98g of H_(2)SO_(4)=1"gram molecules"` `4.9"g of "H_(2)SO_(4)=(4.9)/(98)=0.05"gram molecule"` (b) `underset("1mol")(H_(2)O) =underset("2 gram aroms")(2H)+underset(1"grams atom")(O)` 1 mole of water `(H_(2)O)` has H atoms=2grams 0.15 moles of water `(H_(2)O)` has H atoms=`(2xx0.15)=0.30grams` `=0.30xx6.022xx10^(23)"atoms"` `=1.81xx10^(23)"atoms"` 1 mole of water `(H_(2)O)` has O atoms =1 gram atoms 0.015 moles of water `(H_(2)O)` has O atoms=0.15gram atoms `=0.15xx6.022xx10^(23)=9.03xx10^(23)"atoms"` |
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| 75. |
A gold sample contains 90%` of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold? |
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Answer» Mass of pure gold in one gram of the sample =`(1xx90)/(100)=0.9g` No. of moles of gold present`=("Mass of gold")/("Molar atomic mass")=((0.9g))/((0197"g mol"^(-1)))=0.046"mol"` One mole of gold contains atoms=`N_(A)=6.022xx10^(23)` 0.046 mole of gold conatins atoms=`6.022xx10^(23)xx0.046=2.77xx10^(21)` |
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| 76. |
Sample A contains one gram molecules of oxygen molecules and sample B contains one mole of oxygen molecules. What is the ratio of the number of molecules in both the sample? |
| Answer» One gram molecules and one gram mole contain the same number of molecules (6-022 x 1025). Therefore,the ratio is 1: 1. | |
| 77. |
Show by means of calculations that 5 moles of `CO_(2)` and `5` moles of `H_(2)O` do not have the same mass. How much is the difference in their masses ? |
| Answer» Correct Answer - `5` moles of `CO_(2) = 5 xx 44 = 220 g ;` 5 moles of `H_(2)O = 5 xx 18 = 90 g` ; Difference `= 130 g` | |
| 78. |
In which one of the following cases the number of hydrogen atoms is more ? Two moles of `HCl` or One mole of `NH_(3)` |
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Answer» (i) Two moles of HCl can be written as `2HCl`. We can see that the two moles of `HCl` contain 2 moles of H atoms (or hydrogen atoms). (ii) One mole of `NH_(3)` contains 3 moles of H atoms (or hydrogen atoms). Now, two moles of HCl contains 2 moles of `NH_(3)` contains 3 moles of hydrogen atoms. It is obvious that 1 mole of `NH_(3)` contians more hydrogen atoms. |
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| 79. |
In a reaction 4.0g of sodium carbonate were reacted with 10g of hydrochloric acid solution. The product was a mixture of 2.5g of carbon dioxide and 11.5g of sodium chloride solution. Is this data in agreement with the law of conservation of mass? |
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Answer» The chemical reaction to product is `"sodium chloride"+"hydrochloric acid" to "sodium chloride solution"+"carbon dioxide"` `"Mass of reactants"=(4.0+10.0)=14.0g` `"Mass of products"=(11.5+2.5)=14.0g` |
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| 80. |
Calculate the molecular masses of the following : (a) Hydrogen, `H_(2)` , (b) Oxygen, `O_(2)`, (c ) Chlorine, `Cl_(2)` , (d) Ammonia, `NH_(3)` , (e ) Carbon dioxide, `CO_(2)` (Atomic masses : `H = 1 u, O = 16 u , Cl = 35.5 u , N = 14 u , C = 12 u`) |
| Answer» Correct Answer - (a) `2 u` , (b) `32 u` , (c ) `71 u` , (d) `17 u` , (e ) `44 u`, | |
| 81. |
Gram molecular mass of ammonia `(NH_(3))` is 17 g Is it correct to regard it as formula unit mass also? |
| Answer» No, it is not correct. Ammonia exists in molecular torm and is not an ionic compound made up of cation and anion. Therefore, it cannot have formula unit mass. It has only molecular mass. | |
| 82. |
Which has more number of atoms? 100g of `N_(2)` or 100g of `NH_(3)` |
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Answer» Gram molar mass of `N_(2)=2xx14=28g` 28g of `N_(2)` have nitrogen atoms=`2xxN_(A)` `"100 g "N_(2)"have nitrogen atoms"=2xxN_(A)xx((100g))/((28g))=7.143xxN_(A)` `=7.143xx6.022xx10^(23)=4.3xx10^(24)"atoms"` (b) Gram molar mass of `NH_(3)=14+xx1=17g` `"17 g of "NH_(3) "have atoms"=4xxN_(A)` `"100 g of "NH_(3) "have atoms"=4xxN_(A)xx((100g))/((17g))=2.53xxN_(A)` `=23.53xx6.022xx10^(23)=1.42xx10^(25)"atoms"` 100g of `NH_(3)` have more number of atoms. |
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| 83. |
Which has more atoms? (a)10g of nitrogen `(N_(2))`? 10g of ammonia `(NH_(3))` |
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Answer» Gram molar mas of `N_(2)=2xx14=28g` `28"g of "N_(2)"have nitrogen atoms"=2xxN_(A)` `10"g of "N_(2)"have nitrogen atoms"=2xxN_(A)xx((10g))/((28g))=0.7143xxN_(A)=0.7143xx6.022xx10^(23)` `=4.3xx10^(23)"atoms"` (b) `"Molar mass of "NH_(3)=14xx3xx1=17g` `17"g of "NH_(3)"have atoms"=4xxN_(A)` `10"g of "NH_(3)"have atoms"=4xxN_(A)xx(10g)/(17g)=2.353xxN_(A)` `=2.353xx6.022xx10^(23)=14.16xx10^(23)"atoms"` This shows that 10g of `NH_(3)` have more number of atoms that 10g of `N_(2)`. |
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| 84. |
Name the element water is made of. What are the valencies of the element ? Work out the chemical formula of water. |
| Answer» Correct Answer - `H_(2)O` | |
| 85. |
Calculate the mass of the following: (i) 0.5 mole of `N_(2)` gas (mass from mole of molecule) (ii) 0.5 mole of N atoms (mass from mole of atom) (iii) `3.011 × 10_(23)` number of N atoms (mass from number) (iv) `6.022 × 10_(23)` number of `N_(2)` molecules (mass from number) |
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Answer» (i) mass = molar mass × number of moles `⇒ m = M × n = 28 × 0.5 = 14 g` (ii) mass = molar mass × number of moles `⇒ m = M × n = 14 × 0.5 = 7 g` (iii) The number of moles, n = `("given number of particles")/("Avogadro number") = (N)/(N_(0))` `=(3.011xx10^(23))/(6.022xx10^(23))` `impliesm = M xx n = 14 xx (3.011 xx 10^(23))/(6.022 xx 10^(23))` `=14xx0.5 = 7 g` `n = (N)/(N_(0))` `implies m = M xx (N)/(N_(0))= 28 xx(6.022xx10^(23))/(6.022xx10^(23))` `= 28 xx 1= 28 g` |
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| 86. |
Work out the formula for sulphur dioxide. (Valencies : `S = 4 , O = 2`) |
| Answer» Correct Answer - `SO_(2)` | |
| 87. |
Calculate the number of particles in each of the following: (i) 46 g of Na atoms (number from mass) (ii) `8 g O_(2)` molecules (number of molecules from mass) (iii) 0.1 mole of carbon atoms (number from given moles) |
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Answer» (i) The number of atoms `= ("given mass")/("molar mass")xx` Avogadro number `implies N= (m)/(M) xx N_(0)` `implies N = (46)/(23)xx6.022 xx 10^(23)` `implies N=12.044xx10^(23)` (ii) The number of molecules ⇒`("given mass")/("molar mass")xx` Avogadro number `implies N= (m)/(M)xxN_(0)` atomic mass of oxygen = 16 u `∴` molar mass of `O_(2)` molecules `=16xx2=32g` `implies N = (8)/(32)xx6. 022xx10^(23)` `impliesN=1.5055xx10^(23)` `~= 1.51 xx 10^(23)` (iii) The number of particles (atom) = number of moles of particles `×` Avogadro number `N=nxxN_(0)=0.1 xx 6.022 xx 10^(23)` `=6.022xx10^(22)` |
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| 88. |
The mass of one molecule of a substance is `4.65xx10^(-23)g`. What is its molecular mass? What could be substance be? (b)Which have more molecules? 10g of sulphur dioxide `(SO_(2))` or 10g of oxygen `(O_(2))`? |
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Answer» (a) Mass of one molecules of substance =`4.65xx10^(-23g)` `"Mass of "6.022xx10^(23)"molecules of substance"=6.022xx10^(23)xx(4.65xx10^(-23))=28g` The substances can be carbon monoxide (CO) with molecular mass=12+16=28u or 28g (b) `"Molar mass of "SO_(2)=32h+2xx16g=64g` `"64g of sulphur dioxide represent molecules"=6.022xx10^(23)` `"10g of sulphur dioxide represent molecules"=(6.022xx10^(23)xx(10g))/((64g))=9.40xx10^(22)` `"Molar mass of oxygen "(O_(2))=32g ` 32g of oxygen represent molecules =`6.022xx10^(23)` `"10g of oxygen represent molecules"=(6.022xx10^(23)xx(10g))/((32g))=1.88xx10^(23)` |
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| 89. |
Calculate the mass of `1.2044 xx 10^(23)` molecules of `O_(2) ("Atomic mass of O"=16u)`. |
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Answer» `6.022 xx 10^(23) (N_(A))` molecular of `O_(2)` have mass = 32 u `1.2044 xx 10^(23)` molecular of `O_(2)` have mass `=((32u) xx (1.2044 xx 10^(23)))/((6.022 xx 10^(23))) = 6.4 u` |
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| 90. |
The mass of a single atom of an element X is `2.65 xx 10^(-23) g`. What is its atomic mass ? What could this element be ? |
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Answer» The atomic mass of element is numerically euqal to the mass of 1 mole of its atoms. Since 1 mole of atoms is `6.022 xx 10^(23)` atoms, so it means that the atomic mass of an element is numerically equal to the mass of its `6.022 xx 10^(23)` atoms. Now, `1` atom of element X has mass `= 2.65 xx 10^(-23) g` So, `6.022 xx 10^(23)` atoms of element X have mass `= 2.65 xx 10^(-23) xx 6.022 xx 10^(23)` `= 15.96 g` `= 16 g` Thus, the atomic mass of the element X is 16 u. The element of atomic mass 16 u is oxygen, having the symbol O. |
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| 91. |
An element X forms the following compounds with hydrogen, carbon and oxygen ? `H_(2)X, CX_(2), XO_(2), XO_(3)` |
| Answer» Correct Answer - `2, 4` and `6` | |
| 92. |
If the valency of carbon is 4 and that of sulphur is 2, work out the formula of the compound formed by the combination of carbon with sulphur. What is the name of this compound ? |
| Answer» Correct Answer - `CS_(2)`; carbon disulphide | |
| 93. |
Which of the following statements is not ture about atoms?A. Atoms are not able to exist independentlyB. Atoms are basic units from which molecules and ions are formedC. Atoms are always neutral is natureD. Atoms aggregate in large to form the matter that we can see feel or touch |
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Answer» Correct Answer - A (a) Atoms of inert gas elements can exist independently. However, atoms of all other elements cannot exist independently. |
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| 94. |
Which of the following correctly represents 360 g of water? (i) 2 moles of `H_(2)O` ii) 20 moles of water. iii) `6.022 xx 10^(23)` molecules of water. iv) `1.2044 xx 10^(25)` molecules of water.A. (i)B. I and ivC. ii and iiiD. ii and iv |
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Answer» Correct Answer - D (d) (ii) `("Mass of water")/("Molar mass of water")` `=((360 g))/((18 g"mol"^(-1)))=20 ` mol (iv) 18 g of water represent molecules `=6.022xx10^(23)` 360 g of water represent molecules `=((6.022xx10^(23)))/((18g)) xx(360g)` `=6.022xx10^(23) xx20 =1.2044xx10^(25)` |
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| 95. |
What is a group of `6.022 xx 10^(23)` particles known as ? |
| Answer» Correct Answer - One mole | |
| 96. |
The mass of one atom of an element `X` is `2.0 xx 10^(-23) g`. (i) Calculate the atomic mass of element X. (ii) What could element X be ? |
| Answer» Correct Answer - A::B::C | |
| 97. |
If the aluminium salt of an anion X is `Al_(2)X_(3)`, what is the valency of X ? What will be the formula of the magnesium salt of X ? |
| Answer» Correct Answer - `2 ; MgX` | |
| 98. |
What name is given to the number `6.022 xx 10^(23)` ? |
| Answer» Correct Answer - Avogadro number | |
| 99. |
Calculate formula unit mass of `Al_(2)(SO_(4))_(3)` |
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Answer» Formula unit mass of `Al_(2)(SO_(4))_(3)`, (Aluminium sulphate) `=(2 xx "Atomic mass of Al")+(3 xx "Atomic mass of S")+(12 xx "Atomic mass of O")` `=(2xx 27)+(3 xx 32)+(12 xx 16)` `=(54+96+192)=342u`. |
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| 100. |
Name the compound `Al_(2)(SO_(4))_(3)`, and mention the ions present in it. |
| Answer» The compound is called aluminium sulphate, cation : `Al^(3+), "anion": (SO_(4))^(2-)` | |