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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.
| 101. |
The number of mole of N - atom in `18.066 x 10 ^(23)` nitrogen atoms isA. 1 molB. 2 molC. 3 molD. 4 mol |
| Answer» Correct Answer - C | |
| 102. |
`HNO_(3)` oxidies `NH_(4)^(+)` ions to nitrogen and itself gets reduced to `NO_(2)`. The moles of `HNO_(3)` required by 1 mole of `(NH_(4))_(2)SO_(4)` is:A. (a)`4`B. (b)`5`C. (c )`6`D. (d)`2` |
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Answer» Correct Answer - C `HNO_(3)+NH_(4)^(+) rarr N_(2)+NO_(2)` V.F. of `HNO_(3)=(5-4)=1` `V.F. of NH_(4)^(+)=[0-(-3)]=3` so molar ratio of `HNO_(3)` and `NH_(4)^(+)` is `3:1`. 1 mole `(NH_(4))_(2)SO_(4)` is found to contain 2 mole of `NH_(4)^(+)` So, required moles of `HNO_(3) is 3xx2=6` moles. |
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| 103. |
Calculate mass of one molecule of methane `(CH_(4))` |
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Answer» Mass of `6.022xx10^(23)` molecule of `CH_(4)` Molar mass of `CH_(4)=16 g` and we know mass of 1 molecules `=("Molar mass")/(N)` `=(16)/(6.022xx10^(23))=2.65xx10^(-23)`g |
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| 104. |
Calculate the number of atoms in 0.5 mole atoms of nitrogen. |
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Answer» 1 mole atoms of nitrogen `= 6.022xx10^(23)` atoms `therefore` 0.5 mole atoms of nitrogen `=6.022xx10^(23)xx0.5=3.01xx10^(23)` atoms. |
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| 105. |
At one time, there was a atomic mass scale on the assignment of the value 16.0000 to naturally occuring oxygen. The atomic mass of silver on this scale will be (atomic masses of silver and oxygen on the present scale are 107.868 and 15.9994 respectively).A. `108.000`B. `107.872`C. `108.012`D. `107.909` |
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Answer» Correct Answer - B `("At. Mass of silver on old scale")/("At. mass of" .^(16)O "on old scale")` `=("At. Mass of silver on new scale")/("At. Mass of" .^(16)O "on new scale")` `("At. Mass of silver on old scale")/(16.000)=(107.868)/(15.994)` At. Mass of silver on old scale `=(107.868xx16.000)/(15.9994)=107.872` |
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| 106. |
Eq. mass of `A_(x)B_(y)` isA. `x xx"Eq. mass of A+y"xx"Eq. mass of B"`B. `yxx"Eq. mass of A+x"xx"Eq. mass of B"`C. Eq. mass of A + Eq. mass of BD. None of these |
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Answer» Correct Answer - C Eq. mass of `A_(x)B_(y)=` Eq. mass of A + Eq. mass of B |
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| 107. |
Out of atomic mass, mass number and atomic number, the physical quantities which are not fundamental physical constants is/areA. Atomic massB. Atomic mass and mass numberC. Mass numberD. Atomic number |
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Answer» Correct Answer - A Atomic mass in not a fundamental physical contant. |
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| 108. |
The prefix `10^(18)` isA. gigaB. exaC. kiloD. nano |
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Answer» Correct Answer - B exa `= 10^(18)` |
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| 109. |
Identify the following as homogeneous and heterogeneous mixtures. (i) Sugar dissolved in water. (ii) Oil and water |
| Answer» Sugar solution is homogeneous and has even distribution, oil and water are immiscible therefore heterogeneous mixture. | |
| 110. |
Classify the following substances into element, compound and mixture. (i) Dry ice. (ii) Sodium chloride. |
| Answer» Dry ice (solid `CO_(2)`) and sodium chloride both are compounds. | |
| 111. |
Classify the following substances into element, compound and mixture. (i) Brass (ii) Distilled water |
| Answer» Brass contains copper and zinc is a homogeneous mixture. Distilled water is a compound. | |
| 112. |
Identify the following as homogeneous and heterogeneous mixture. (i) Aerated drinks (ii) Brass |
| Answer» Brass is homogeneous mixture because constituents are evenly mixed whereas aerated drinks have liquid and separate from each other, when opened, so a heterogeneous mixtures. | |
| 113. |
In which case is the number of molecules of water maximum?A. (a)18 mL of waterB. (b)0.18 g of waterC. (c )0.00224 L of water vapour at 1 atm and 273 KD. (d)`10^(-3)` mol of water |
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Answer» Correct Answer - d (a) `18 mL` water As `d_(H_(2)O)=1 g//mL` So `W_(H_(2)O)=18 g` `n_(H_(2)O)=18/18=1` molecules `=1xxN_(A)` (b) `0.18` g of water `n_(H_(2)O)=0.18/18=0.01` `("molecules")_(H_(2)O)=0.01 xxN_(A)` (c ) `(V_(H_(2)O(g)))_(STP)=0.00224 L` `n_(H_(2)O)=V/22.4=0.00224/22.4=0.0001` molecules`=0.0001xxNA` (d) `n_(H2O)=10^(-3)` `("molecules")_(H2O)=10^(-3)xxN_(A)` |
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| 114. |
Classify the following substances into element, compound and mixture. (i) Smoke (ii) Mercury. |
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Answer» (i) Smoke is a heterogeneous mixture. (ii) Mercury is an element. |
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| 115. |
In which case is the number of molecules of water maximum?A. 18 mL of waterB. 0.18g of waterC. 0.00224 L of water vapours at 1 atm and 273 KD. `10^(-3)` mol of water |
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Answer» Correct Answer - A (1) 18 mL water As `d_(H_(2)O)=1g//mL" "`So `W_(H_(2)O)=18 g` `n_(H_(2)O)=18/18=1` molecules `=1xxN_(A)` (2) 0.18 g of water `n_(H_(2)O)=0.18/18=0.01` (molecules) `H_(2)O=0.01xxN_(A)` (3) `(V_(H_(2)O(g)))_(STP)=0.00224 L` `n_(H_(2)O)=V/22.4=0.00224/22.4=0.0001` Molecules `=0.0001xxN_(A)` (4) `n_(H_(2)O)=10^(-3)` `("molecules")_(H_(2)O)=10^(-3)xxN_(A)` |
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| 116. |
In a closed vessel, 5 moles of `A_(2)(g)` and 7 moles of `B_(2)` (g) are reacted in the following maner, `A_(2)(g)+(3B_(2)(g)to2AB_(3)(g)` What is the total number of moles of gases present in the container at the end of the reaction?A. `22//3`B. `7//3`C. `14//3`D. `8//3` |
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Answer» Correct Answer - B `underset("Initial 5 mol")(A_(2)) + underset(" 7 mol")(3B_(2)) rarr underset(0)(2 AB_(3))` By reactionn 1 mol `A_(2)` requires 3 mol `B_(2)` Hence, 7 mol `B_(2)` will reacts with `7/3` moles `A_(2)` `A_(2) ` left`= 5 - 7/3 = 8//3 ` mol `A_(2)` (g) 1 mol `A_(2)` produces 2 mol `AB_(3)` Hence, `7/3` mol ` A_(2)` will produce `7/3xx 2` mol `AB_(3)` Total mol of gases in vessel `= 8/3 mmol `A_(2) (g) + 14/3` mol `AB_(3)` (g) `= 22/3` mol |
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| 117. |
Atomicity of mercury vapour isA. `1`B. `2`C. `3`D. `4` |
| Answer» Correct Answer - A | |
| 118. |
The total number of protons in 10g of calcium carbonate is `(N_(0) = 6.023xx10^(23))`A. `1.5057xx10^(24)`B. `2.0478xx10^(24)`C. `3.0115xx10^(24)`D. `4.0956xx10^(24)` |
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Answer» Correct Answer - C Protons in 1 mole `CaCO_(3)` = atomic number Ca + atomic number C+3`xx`atomic number O `20+6+24 = 50` moles Protons in 10 g `= (10)/(100)xx50xx6.02xx10^(23)` `=3.01xx10^(24)` |
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| 119. |
Assertion : Atomicity of oxygen is 2. Reason : 1 mole of an element contains `6.023xx10^(23)` atoms.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B No. of atoms present in a molecules of a gaseous element is called atomicity. For example, `O_(2)` has two atoms and hence its atomicity is 2. |
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| 120. |
One mole of fluorine is reacted with two moles of hot and concentrated KOH. The product formed are KF, `H_(2)O` and `O_(2)`. The molar ratio of `KF, H_(2)O` and `O_(2)` is respectivelyA. `1:1:2`B. `2:1:0.5`C. `1:2:1`D. `2:1:2` |
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Answer» Correct Answer - B `F_(2)+2KOHto2KF+H_(2)O+1//2O_(2)` `:.` Molar ratio of `KF:H_(2):O_(2)` from balanced chemical equation is 2:1:0.5(1/2) |
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| 121. |
If isobutane and n-butane are present in a gas, then how much oxygen should be required for complete combustion of 5 kg of this gasA. 17.9 kgB. 9 kgC. 27 kgD. 1.8 kg |
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Answer» Correct Answer - A Isobutane and n-butane `[C_(4)H_(10)]` have same molecular formula, `C_(4)H_(10)+13/2 O_(2) rarr 4 CO_(2)+5 H_(2)O` For 58 gm of `C_(4)H_(10)` 208 gm `O_(2)` is required then for 5 kg of `C_(4)H_(10)" "O_(2)=(5xx208)/58=17.9 kg` |
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| 122. |
Which of the following correctly represents 180 g of water ? 5 moles of water (ii) 10 moles of water (iii) `6.023xx10^23` molecules of water (iv) `6.023xx10^24` molecules of waterA. (i) and (ii)B. (i) and (iv)C. (ii) and (iv)D. (ii) and (iii) |
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Answer» 18g of `H_2O`=1 mol 180g of `H_2O` = 10 mol 18g of `H_2O=6.023xx10^23 " molecules of " H_2O` 180g of `H_2O = (6.023xx10^23)/(18)xx180` `=6.023xx10^24 " molecules of " H_2O` |
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| 123. |
Assertion: Equivalent mass of ozone in the change `O_(3) rarr O_(2) is 8`. Reason: 1 moles of `O_(3)` on decomposition gives `3//2` mole of `O_(2)`.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. |
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Answer» Correct Answer - B `2O_(3) rarr 3O_(2), 2 "mole" O_(3) -= 3 "moles" O_(2)=3xx4 eq. O_(2)` `:. E_(O_(3))=M/6=48/6=8 EO`. |
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| 124. |
In a reaction container, 100g of hydrogen and 100 g of `Cl_2` are mixed for the formation of HCl gas. What is the limiting reagent and how much HCl is formed in the reaction ?A. `H_2` is limiting reagent and 36.5 g of HCl are formed.B. `Cl_2` is limiting reagent and 102.8 g of HCl are formed.C. `H_2` is limiting reagent and 142 g of HCl are formed.D. `Cl_2` is limiting reagent and 73 g of HCl are formed. |
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Answer» `underset(2g)(H_2)+underset(71g)(Cl_2) to underset(73g)(2HCl)` 2g of `H_2` reacts with 71 g of `Cl_2` 100 g of `H_2` reacts with `71/2xx100=3550g " of " Cl_2` Hence, `Cl_2` is the limiting reagent. 71g of `Cl_2` produces 73 g of HCl 100g of `Cl_2` will produce `73/71xx100=102.8g` of HCl |
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| 125. |
If 40g of `CaCO_3` is treated with 40g of HCl, which of the reactants will acts as limiting reagent?A. `CaCO_3`B. HClC. Both (a) and (b)D. None of these |
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Answer» `underset("1 mole" (100g))(CaCO_3)+underset("2 moles" (73g))(2HCl) to CaCl_2+H_2O+CO_2` 100g of `CaCO_3` reacts with 73g of HCl 40g of `CaCO_3 " will reacts with " 73/100xx40=29.2g` of HCl Since `CaCO_3` is completely consumed and some amount (40-29.2=10.8g) of HCl remains unreacted and hence, `CaCO_3` is limiting reagent. |
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| 126. |
How much oxygen is required for complete combustion of 560 g of ethene?A. 6.4 kgB. 1.92 KgC. 2.8 kgD. 9.6 kg |
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Answer» `underset(28g)(C_2H_4)+underset(3xx32=96g)(3O_2) to 2CO_2 + 2H_2O` 28 g of `C_2H_4` requires 96g of `O_2` 560 g of `C_2H_4` requires `96/28xx560` =1920g or 1.92kg of `O_2` |
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| 127. |
How many moles of oxygen gas can be produced during electricity decomposition of 180 g of water ?A. 2.5 molesB. 5 molesC. 10 molesD. 7 moles |
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Answer» `underset(2xx18=36g)(2H_2O) to 2H_2 + O_2` 36 g of water produces 1 mole of `O_2` gas 180 g of water will produce 180/36 = 5 moles of `O_2` gas. |
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| 128. |
When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes fromA. 7 to 4B. 6 to 4C. 7 to 2D. 4 to 2 |
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Answer» Correct Answer - C `{:(" COOH"),(5" | " +2 Koverset(+7)(M)nO_(4)+3H_(2)SO_(4) rarr K_(2)SO_(4)+2 overset(+2)(M)nSO_(4)+10 CO_(2) +8 H_(2)O),(" COOH"):}` In this reaction oxidation state of Mn change from +7 to +2. |
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| 129. |
The oxidation states of sulphur in the anions `SO_(3)^(2-), S_(2)O_(4)^(2-)`, and `S_(2)O_(6)^(2-)` follow the orderA. `S_(2)O_(6)^(2-) lt S_(2)O_(4)^(2) lt SO_(3)^(2-)`B. `S_(2)O_(4)^(2-) lt SO_(3)^(2-) lt S_(2)O_(6)^(2-)`C. `SO_(3)^(2-) lt S_(2)O_(4)^(2-) lt S_(2)O_(6)^(2-)`D. `S_(2)O_(4)^(2) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)` |
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Answer» Correct Answer - B `S_(2)O_(4)^(2-) lt SO_(2)^(2-) lt S_(2)O_(6)^(2-)` Oxidation state of sulphur in `S_(2)O_(4)^(2-)=+3` Oxidation state of sulphur in `SO_(3)^(2-) =+4` Oxidation state of sulphur in `S_(2)O_(6)^(2-) = +5`. |
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| 130. |
Which of the following chemical reactions depicts the oxidizing behaviour of `H_(2)SO_(4)` ?A. `2HI+H_(2)SO_(4) rarr I_(2)+SO_(2)+2H_(2)O`B. `Ca(OH)_(2)+H_(2)SO_(4) rarr CaSO_(4)+2H_(2)O`C. `NaCl+H_(2)SO_(4) rarr NaHSO_(4)+HCl`D. `2 PCl_(5)+H_(2)SO_(4) rarr 2POCl_(3)+2HCl +SO_(2)Cl_(2)` |
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Answer» Correct Answer - A `2 overset(-1)(HI)+H_(2)overset(+6)(S)O_(4) rarr overset(0)(I_(2))+overset(+4)(SO_(2))+2H_(2)O`. |
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| 131. |
What mass of oxygen gas is required to burn completely `2.5 mol` of methane? `underset(underset("reactant")("one"))("Moles of ")rarr underset(underset("reactant")("another"))("Mass of")` Stragetgy : Use the balanced equation `{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}` to find the relationship among moles and grams of reactants : `underset(CH_(4))("Mole of ")rarr underset(O_(2))("Moles of")rarr underset(O_(2))("grams of")` |
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Answer» We can solve the problem in three steps. Step1: Mass `CH_(4) rarr mol CH_(4)` `? "mol" CH_(4) = 40.0g CH_(4) xx (1 mol CH_(4))/(16.0g CH_(4)) (n = ("Mass")/("Molar mass"))` `= 2.50 "mol" CH_(4)` Step 2 : `"mol" CH_(4) rarr mol O_(2)` `? "mol" O_(2) = 2.50 "mol" CH_(4) xx (2 "mol" O_(2))/(1 "mol" CH_(4))` (Balanced equation used) `= 5.00 "mol" O_(2)` Step 3 : mol `O_(2) rarr` mass `O_(2)` `? g O_(2) = 5.00 "mol" O_(2) xx (32.0 g O_(2))/(1 mol O_(2)) ("Mass = n xx MM")` `= 160g O_(2)` Alternaty all these steps can be combined into one setup as follows : `underset(CH_(4))("g of")rarrunderset(CH_(4))("mol of")rarr underset(O_(2))("mol of")rarr underset(O_(2))("g of")` `? g O_(2) = 40.0g CH_(4) xx (1 "mol" CH_(4))/(16.0g CH_(4)) xx (2 "mol" O_(2))/(1 "mol" CH_(4))` `xx (32.0 g O_(2))/("1 mol" O_(2))` `= 160g O_(2)` |
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| 132. |
What mass of oxygen gas is required to burn completely `2.5 mol` of method? `underset(underset("reactant")("one"))("Moles of ")rarr underset(underset("reactant")("another"))("Mass of")` Stragetgy : Use the balanced equation `{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}` to find the relationship among moles and grams of reactants : `underset(CH_(4))("Mole of ")rarr underset(O_(2))("Moles of")rarr underset(O_(2))("grams of")` |
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Answer» Accroding to the equcation, 2 mol of `O_(2)` is required for every 1 mol of `CH_(4)`. Thus, we need `2xx2.5(=5.0) mol O_(2)`. Mass of 1 mol `O_(2)` is `32g`. Thus, mass of `5 mol O_(2)` is `5xx32 = 160 g`. Alternatively, `? g O_(2) = 2.5 "mol" CH_(4) xx (2 "mol" O_(2))/(1 "mol" CH_(4)) xx (32.0g O_(2))/("1 mol" O_(2))` `= 160 g O_(2)` |
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| 133. |
Two metallic oxides contain `27.6%` and `30%` oxygen, respectively. If the formula of the second oxide is `M_(2)O_(3)`, that of the first will beA. `M_(3) O_(4)`B. `M_(2) O_(5)`C. `MO_(3)`D. `M_(2) O` |
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Answer» Correct Answer - A In `M_(2) O_(3)`, 30 parts of `O` combine with 70 (parts of `M`. Thus, 70 parts of `M` in `M_(2)O` correspond to 2 atoms of `M`. In first oxide, `72.4` parts will correspond to `(2)/(70) xx 74.7 = 2.07 M` atoms. 30 parts of `O` in `M_(2) O_(3)` corrrespond to 3 atom of `O`. In first oxide, 27.6 parts of `O` will correspond to `(3)/(30) xx 27.6 = 2.76 O` atoms. Ratio of `M` to `O` in the first oxide `= 2.07 : 2.76 = (2.07)/(2.07) = (2.76)/(2.07) = 1 : 1 : 33 = 3 : 4` |
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| 134. |
Iron forms two oxides, in first oxide `56` grams. Iron is found to be combined with `16` grams oxygen and in second oxide `112` grams iron is found to be combined with `48` grams oxygen. This data satisfy the law ofA. (a)Conservation of massB. (b)Reciprocal proportionC. (c )Multiple proportionD. (d)Combining volume |
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Answer» Correct Answer - C For fix mass of Fe, i.e., `56` parts, the masses of oxygen in these two oxides are `16` and `48` respectively, simple ratio is `1:3` so it is law of multiple proportion |
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| 135. |
Zinc`(Zn)` a silvbery metal is used to from brass (with copper) and to plate iron to prevent corrosion. How many grams of `Zn` are there in `0.25 mol` of `Za`?A. `16g`B. `13g`C. `15g`D. `20g` |
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Answer» Correct Answer - A Molar mass of `Zn` in `65.39g`. `0.25 mol Za xx (65.39g zn)/(1 mol Zn) = 16g Zn` |
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| 136. |
If `10^(21)` molecules are removed from 200 mg of `CO_(2)`, then maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed isA. `2.88xx10^(-3)`B. `1.66xx10^(-3)`C. `4.54xx10^(-3)`D. `1.66xx10^(-2)` |
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Answer» Correct Answer - A `10^(21)` molecules of `CO_(2)` `(10^(21))/(6.02xx10^(23))`moles `=1.66xx10^(-3)` moles 200 mg of `CO_(2)` left `=(200xx10^(-3))/(44)`moles `=4.54xx10^(-3)` moles Moles of `CO_(2)` left `=(4.54-1.66)xx10^(-3)` `=2.88xx10^(-3)` moles |
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| 137. |
If `3*01xx10^(20)` molecules are removed from 98 mg of `H_(2)SO_(4)`, then the number of moles of `H_(2)SO_(4)` left areA. `0*1xx10^(-3)`B. `0*5xx10^(-3)`C. `1*66xx10^(-3)`D. `9*95xx10^(-2)` |
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Answer» Correct Answer - B Moles of `H_(2)SO_(4)` removed `= (3*01xx10^(20))/(6*02xx10^(23))=0*5xx10^(-3)` Total moles of `H_(2)SO_(4) = (98xx10^(-3))/(98) = 1xx10^(-3)` Moles of `H_(2)SO_(4)` left `=1xx10^(-3)-0*5xx10^(-3)` `=0*5xx10^(-3)` moles |
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| 138. |
The number of equivalents of `N_(2)S_(2)O_(3)` required for the volumetric estimation of one equivalent of `Cu^(2+)` isA. 1B. 2C. `3//2`D. 3 |
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Answer» Correct Answer - B `Cu^(2+)+2I^(-) rarr CuI_(2)" "2 CuI_(2) rarr Cu_(2)I_(2)+I_(2)` `I_(2)+2NA_(2)S_(2)O_(3) rarr 2NaI+Na_(2)S_(4)O_(6)` `Cu^(2+) equiv 2 Na_(2)S_(2)O_(3)` |
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| 139. |
Assertion: 1 equivalent of `H_(2)SO_(4)` contains 1 equivalent of `H, S` and `O` each Reason: A species contains same number of equivalents of its components.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. |
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Answer» Correct Answer - A Equivalent reacts in equal number. |
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| 140. |
The number of g-molecules of oxygen in `6.0xx10^(24)` CO molecules is: `[Take:N_(A)=6xx10^(23)]`A. 10B. 5C. 1D. 0.5 |
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Answer» b) `6.02 xx 10^(24)` CO molecules = 10 moles of CO =10 g atoms of O = 5g molecules of `O_(2)` |
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| 141. |
Reduction of the metal centre in aqueous permanganate ion involvesA. `3 e^(-)` in neutral mediumB. `5 e^(-)` in neutral mediumC. `3 e^(-)` in alkaline mediumD. `5 e^(-)` in acidic medium |
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Answer» Correct Answer - A::D In acidic medium `MnO_(4)^(-) rarr Mn^(2+)` (v.f = 5) In neutral /basic medium `MnO_(4)^(-) rarr MnO_(2)` (v. f. = 3) In strongly basic medium `MnO_(4)^(-) rarr MnO_(4)^(2-)` (v.f. = 1) |
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| 142. |
The order of the oxidation state of the phosphours atom in `H_(3)PO_(2),H_(3)PO_(4),H_(3)PO_(3)` and `H_(4)P_(2)O_(6)` isA. `H_(3)PO_(4) gt H_(3)PO_(2) gt H_(3)PO_(3) gt H_(3)P_(2)O_(6)`B. `H_(3)PO_(4) gt H_(4)P_(2)O_(6) gt H_(3)PO_(3) gt H_(3)PO_(2)`C. `H_(3)PO_(2) gt H_(3)PO_(3) gt H_(4)P_(2)O_(6) gt H_(3)PO_(4)`D. `H_(3)PO_(3) gt H_(3)PO_(2) gt H_(3)PO_(4) gt H_(4)P_(2)O_(6)` |
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Answer» Correct Answer - B Correct order : `{:(H_(3)PO_(4),gt,H_(4)P_(2)O_(6),gt,H_(3)PO_(3),gt,H_(3)PO_(2)),((+5),,(+4),,(+3),,(+1)):}` |
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| 143. |
Mass of 0.1 moleof methane isA. 1gB. 16 gC. 1.6 gD. 0.1 g |
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Answer» Mass of 1 mole of methane `(CH_(4))=16g` Mass of 0.1 mole of methane `=16xx0.1g=1.6g` |
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| 144. |
Calculate the number of `H` atoms in `39.6g` of ammonium sulphate, `(NH_(4))_(2) SO_(4)`.A. `4.41xx10^(24) H` atomsB. `7.86xx10^(24) H` atomsC. `1.44xx10^(24) H` atomsD. `5.72xx10^(24) H` atoms |
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Answer» Correct Answer - C `underset(underset("sulphate")("ammonium"))("Mass of")rarr underset(underset("sulphate")("ammonium"))("Moles of")rarr underset(underset("sulphate")("of ammonium"))("Formula units")rarr underset(underset("atoms")("of H"))("Number")` `36.6g(NH_(4))_(2)SO_(4)xx(1"mol"(NH_(4))_(2)SO_(4))/(132.1g(NH_(4))_(2)SO_(4))` `xx(6.022xx10^(23)"formula units")/(1 "mol"(NH_(4))_(2)SO_(4))xx("8 H atoms")/("1 formula unit")` `=1.44xx10^(24)` H atoms |
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| 145. |
If `13.1g` of `Na_(2) SO_(4) XH_(2)O` contains `6g` of `H_(2)O`, the value of `X` isA. `10`B. `5`C. `3`D. `7` |
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Answer» Correct Answer - D According to formula, 1 mol `Na_(2) SO_(4) (142g)` always combines with `X` moles `H_(2)O (18X g)`. According to data, mass of `Na_(2) SO_(4) = (13.1 - 6)g = 7.1g`. In a compound the ratio of mass is always fixed. Thus, `(.^(m)Na_(2)SO_(4))/(.^(m)H_(2)O) = (142g)/(18Xg) = (7.1)/(6g)` `:. X = ((42xx6))/((18xx7.1)) = (852)/(127.8) = 6.66 ~~ 7` |
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| 146. |
Which of the following samples contains the largest number of atoms ? .(a). 1g of Ni(s)(b). 1g of Ca(s)(c). 1g of `N_(2)(g)`(d). 1g of B(s)A. 1 g of Ni(s)B. 1 g of Ca(s)C. 1 g of `N_(2)` (g)D. 1 g of B (s) |
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Answer» Correct Answer - D Lesser the atomic mass, more is the number of atoms per gram. At mass of Ni, Ca, N and B are 58.7, 40, 14 and 10.8 respectively. |
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| 147. |
Emprical formula: Write the emprical formulas for the follwing molecule compounds: (i) dinitrogen tetoxide `(N_(2) O_(4))`, (ii) acterylene `(C_(2) H_(12) O_(6))`. Strategy: Divide all the subscipts, if possible by a sutiable number so that the subscips are convered to the smllest whole numbers. |
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Answer» (i) Dividing the subsrips by 2 gives the emprical formula `NO_(2)`, (ii) Dividing the the suscrips by 2 gives the emprical fromula `CH`. (iii) As there is no number by which both the subscrips 2 and are divisible, the empirical formula for indine pentoxide the same as molecular formula. (iv) Dividing the subscrips by 6 gives the emprical formula `CH_(2) O` |
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| 148. |
32 g `O_(2)`, 2g `H_(2)` and 28 g `N_(2)` at S.T.P. occupy separately a volume ofA. 1LB. 2LC. 22.4 LD. 2.24 L |
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Answer» Correct Answer - C One mole of every gas at S.T.P. occupies 22.4 L |
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| 149. |
Number of hydrogen atoms in 36 ml of `H_(2)O` at 277 K are x `N_(A)`. X is ________. |
| Answer» Correct Answer - D | |
| 150. |
Assertion A `8.0 g N_(2)H_(4) (M = 32)` has more atoms than `6.0 g H_(2)O`. Reason : `N_(2)H_(4)` has more atoms per molecule than water.A. Both assertion and reason are correct and reason is the correct explanation of the assertion.B. Both assertion and reason are correct but reason is not the correct explanation of assertion.C. Assertion is correct but reason is incorrect.D. Assertion is incorrect but reason is correct. |
| Answer» Correct Answer - B | |