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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.
| 901. |
The total number of electrons in one molecular of carbon dioxide isA. (a)`22`B. (b)`44`C. (c )`66`D. (d)`88` |
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Answer» Correct Answer - A No. of `e^(-)` in `C=6` and in `O=8` :. Total no. of `e^(-)` in `CO_(2)=6+8xx2=22`. |
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| 902. |
The number of significant figures in `38.57 mL` isA. twoB. fourC. infiniteD. three |
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Answer» Correct Answer - B Nonzero digits are always significant |
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| 903. |
If the density of a solu tion is 3.12 g `mL^(-1)` , the mass of 1.5mL solution in significant figures is.........A. 4.7gB. `4680xx10^(-3)g`C. `4.680g`D. `46.80g` |
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Answer» Given that density of solution =`312g mL^(-1)` Volume of solution = 15mL For a solution Mass =volume xx density `=1.5mL xx3.12 g mL^(-1)=4.68g` The digit 1.5has only two significant figures, so the answer must also be limited to two significant figures, So, it is rounded off to reduc e the number of significant figures. Hence, the answer is reported as 4.7g . |
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| 904. |
Which of the following statements about a compound is incorrect?A. A molecule of a compound has atoms of different elementsB. A compound cannot be separated into its constituent elements by physical methods of separationC. A compound retains the physical proepreties of its constituent elementsD. the ratio of atoms of different elements in a compound is fixed. |
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Answer» A compound is a pure substance containing two or more than two elements combined together in a fixed proportion by mass and which can be decomposed into its constituent elements by suitable chemical methods Further, the procperites of a compound are quite different from the properties of constituent elements e.g. water is a compound contianing hydrogen and oxygen combined together in a fixed proportionation. But the properties of water are completely different from its consituents, hydrogen and oxygen. |
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| 905. |
Assertion(A) Combustion of 16g of methane give18 g of water. Reason(R) in the combustion of methane, water is one of the products.A. Both A and R are true and R is the correct explnanation of A.B. A is true but R is false.C. A is false but R is true.D. Both A and R are flase. |
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Answer» Assertion is false but Reson is true. Combustion of 16 g of methane gives 36 g of water `CH_(4)+2O_(2)rarrCo_(2)+2H_(2)O` imol 2 mol =16g =36g |
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| 906. |
1.5g `CdCl_(2)` was formed to contain 0.9g Cd. Calculate the atmic weight of Cd.A. 118B. 112C. 106.5D. 53.25 |
| Answer» Correct Answer - B | |
| 907. |
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. (a)27 gB. (b)54 gC. (c )49.5 gD. (d)31 g |
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Answer» Correct Answer - B `2Al+3/2O_(2)rarr Al_(2)O_(3)` According to equation `3/2` mole of `O_(2)` combines with 2 moles Al. 2 mole `Al=54 g` |
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| 908. |
If 6.3 g of `NaHCO_(3)` are added to 15.0 g of `CH_(3)COOH` solution, the residue is found to weigh 18.0 g. The mass of `CO_(2)` released in the reaction isA. 9.3 gB. 39.3 gC. 3.3 gD. None of these |
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Answer» Correct Answer - C `CH_(3)COOH+NaHCO_(3) to CH_(3)COONa+H_(2)O+CO_(2) uarr` Mass of reactants = Mass of products `6.3 g + 15.0 g= 18.0 g+` Mass of `CO_(2)` Mass of `CO_(2) = 6.3g + 15.0g - 18.0 g = 3.3 g` |
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| 909. |
When 6.4 g of `NaHCO_(3)` is added to a solution of `CH_(3)COOH` weighing 20. It is observed that 4.4 g of `CO_(2)` is released into atmosphere and a residue is left behind Calculate the mass of residue by applying law of conservation of mass. |
| Answer» Correct Answer - 24.0 g | |
| 910. |
consider the following reaction, `Na_(2)CO_(3) + 2HCl to 2NaCl + H_(2)O + CO_(2)` Equivalent weight of `Na_(2)CO_(3)` isA. `M/2`B. MC. 2MD. `M/4` |
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Answer» a) `Na_(2)CO_(3) + 2HCl to 2NaCl = H_(2)O + CO_(2)` In the above reaction, equivalent weight of `Na_(2)CO_(3)` is `M/2` because 2 moles of `Na^(+)` beign transferred per mole of `Na_(2)CO_(3)`. |
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| 911. |
If `1//6`, in place of `1//12`, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one one of a substance will:A. to be a function of the molecular mass of the substanceB. remian unchangedC. increase two foldD. decrease twice |
| Answer» b) Mass of the given amount of substance is a constant quantity, i.e. remains unchanged. | |
| 912. |
Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values. `{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}` Using these data, obtain the correct explanation for the following questions. Among the following, identify the correct statementA. Chloride ion is oxidised by `O_(2)`B. `Fe^(2+)` is oxidised by iodideC. Iodide ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine |
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Answer» Correct Answer - C Calculate the EMF of all the cells. Only the EMF of the cell involving the oxidation of `I^(-)` ion by `Cl_(2)` is +ve. `{:(2I^(-) rarr I_(2)+2e^(-)", "E^(@)=-0.54 V),(Cl_(2)+2e^(-) rarr 2Cl^(-)", "E^(@)=+1.36 V),(bar(Cl_(2)+2I^(-) rarr 2Cl^(-)+I_(2), E_("cell")^(@)+0.82 V)):}` Thus, option (c) is correct. |
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| 913. |
Number of atoms of `He` in `100` atoms of `He` (at.mass 4 amu) isA. 25B. 100C. 50D. `100xx6xx10^(-23)` |
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Answer» Correct Answer - A 1 amu `=1.66xx10^(-24) g` 100 amu `=100xx1.66xx10^(-24)` gm No. of atoms of a substance `=("Wt. of substance")/("No. of gm moles") xx6.023xx10^(23)xx`Atomicity `=(100xx1.66xx10^(-24))/4xx6.023xx10^(23)xx1=25`. |
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| 914. |
Number of atoms of `He` in `100` atoms of `He` (at.mass 4 amu) isA. `100xx6.022xx10^(23)`B. `50`C. `25`D. `100` |
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Answer» Correct Answer - C Nuimber of `He` atoms `= ("Number of atomic mass units")/("Atomic mass of" He)` `= (100 amu)/(4 "amu atom"^(-1)) = 25 He` atoms |
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| 915. |
How many grams of `CaO` are required to neutralise `852 g` of `P_4 O_10` ? Draw the structure of `P_4 O_10`.A. 852gB. 1008gC. 85gD. 7095g |
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Answer» `6CaO+P_4O_10 to 2Ca_3 (PO_4)_2` Mole of `P_4O_10` = molar mass of `P_4O_10` = 284g 852g of `P_4O_10=852/284=3` mol mole of `P_4O_10` reacts with 6 moles of CaCO 3 moles of `P_4O_10` reacts with 18 moles of CaCO Mass of 18 moles of CaO `=18xx56=1008g` |
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| 916. |
Which among the following is the heavist?A. 1 mol of oxygenB. One molecule of sulphur trioxideC. 100 amu of uraniumD. 100 mol of hydrogen |
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Answer» Finding the masses of others `1 mol O_(2) = 32g` 1 molecule `SO_(2) = (64g mol^(-1))/(6.022xx10^(23) "molecules" mol^(-1))` `= 10.62xx10^(-23)g` `= 1.062xx10^(-22)g` 100 amu of uranium `= 100 am u xx (1.66xx10^(-24)g)/(1 am u)` `= 1.66xx10^(-24)g` `10 mol` of hydrogen `= (10 mol) (2g mol^(-1)) = 20g` |
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| 917. |
Common salt obtained from sea water contains `95% NaCl` by mass. The appoximate number of molecules present in `10.0g` of the salt isA. `10^(21)`B. `10^(22)`C. `10^(23)`D. `10^(24)` |
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Answer» Correct Answer - C 58.5 g of NaCl contains `6.023xx10^(23)` molecules. 10 g will contain `1.03xx10^(23)` molecules of NaCl. Since it contain 95 % NaCl, so total number of molecules `=95/100xx1.03xx10^(23)= 0.98xx10^(23) or 10^(23)` |
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| 918. |
How many grams of caustic potash required to completely neutralise `12.6 g HNO_(3)`?A. (a)`22.4 KOH `B. (b)` 1.01 KOH`C. (c )` 6.02 KOH`D. (d)`11.2 KOH ` |
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Answer» Correct Answer - D `HNO_(3)+KOH rarr KNO_(3)+H_(2)O` `12.6/63=0.2 "mole", HNO_(3)-= KOH` `0.2"mole"-= 0.2 "mole"` `0.2xx56=11.2 g`. |
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| 919. |
The oxidation number of `S` in `S_(8),S_(2)F_(2)`, and `H_(2)S`, respectively, areA. `0,+1` and `-2`B. `+2, +1` and `-2`C. `0, +1` and +`2`D. `-2, +1` and `-2` |
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Answer» Correct Answer - A In `S_(8)`, oxidation number of `S` is 0, elemental state. In `S_(2)F_(2), F` is in `-1` oxidation state, hence S is in `-2` oxidation state. In `H_(2)S, H` is in `+1` oxidation state, hence S is in `-2` oxidation state. |
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| 920. |
The oxidation number of sulphur in `H_(2)S_(2)O_(8), H_(2)S_(2), O_(4), H_(2)S_(2)O_(6)` are respectivelyA. `+3, +4, +5`B. `+5, +4, +3`C. `+6, +3, +5`D. `+3, +5, +4` |
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Answer» Correct Answer - C `H_(2)S_(2)O_(8)` : `{:(" O O"),(" || ||"),(HO-S-O-O-S-OH" "S(+VI)),(" || ||"),(" O O"):}` `{:(" O O"),(" || ||"),(H_(2)S_(2)O_(4) : HO-S-S-OH" "S(+III)):}` `{:(" O O"),(" || ||"),(H_(2)S_(2)O_(6) : HO-S-S-OH" "S(+V)),(" || ||"),(" O O"):}` |
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| 921. |
Two grams sulphur is completely burnt in oxygen to form `SO_(2)`. In this reaction, what is the volume (in litres) of oxygen consumed at S.T.P. ?(at mass of S and O are 32 and 16 respectively)A. `(16)/(22.414)`B. `(22.414)/(16)`C. `(22.414)/(32)`D. `(32)/(22.414)` |
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Answer» Correct Answer - B `underset(32g)(S)+underset(22.414L)(O_(2))toSO_(2)` 32 g S required `O_(2)` at S.T.P. = 22414 L 2 g of S require `O_(2)` at S.T.P. `= (22.414)/(32)xxL` `=(22.414)/(16)L` |
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| 922. |
The oxidation number of sulphur in `S_(8), S_(2)F_(2) and H_(2)S` respectively are:A. 0, +1 and -2B. `+2`, +1 and -2C. 0, +1 and +2D. `-2`, +1 and -2 |
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Answer» Correct Answer - A `overset(**)(S)_(8) = 0 rArr overset(**)(S)_(2) F_(2) = +1 rArr H_(2)overset(**)(S) = -2` |
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| 923. |
The oxidation number and covalency of sulphur in the sulphur molecule `(S_(8))` are respectively:A. 0 and 2B. 6 and 8C. 0 and 8D. 6 and 2 |
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Answer» Correct Answer - A The oxidation number of sulphur in the sulphur molecule `(S_(8))` is 0 and 2. |
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| 924. |
Which of the following have been arranged in the decreasing order of oxidation number of sulphur ?A. `Na_(2)S_(4)O_(6) gt H_(2)S_(2)O_(7) gt Na_(2)S_(2)O_(3) gt S_(8)`B. `H_(2)SO_(4) gt SO_(2) gt H_(2)S gt H_(2)S_(2)O_(8)`C. `SO_(2)^(2+) gt SO_(4)^(2-) gt SO_(3)^(2-) gt HSO_(4)^(-)`D. `H_(2)SO_(5) gt H_(2)SO_(3) gt SCl_(2) gt H_(2)S` |
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Answer» Correct Answer - D `{:("Compound","Oxidation state"),(H_(2)SO_(5),+8),(H_(2)SO_(3),+4),(SCl_(2),+2),(H_(2)S,-2):}` |
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| 925. |
The weight of `1xx10^(22)` molecules of `CuSO_(4).5H_(2)O` isA. `41.59 g`B. `415.9 g`C. `4.159 g`D. None of these |
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Answer» Correct Answer - C Mol. Mass of `CuSO_(4).5H_(2)O` `=63+32+16xx4+5xx18=249` `6.02xx10^(23)` molecules of `CuSO_(4).5H_(2)O` has a mass `=((249g)xx1xx10^(22))/(6.02xx10^(23))=4.159g` |
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| 926. |
In the standardization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry, th equivalent weight of `K_(2)Cr_(2)O` isA. `("Molecular weight")//2`B. `("Molecular weight")//6`C. `("Molecular weight")//3`D. Same as molecular weight |
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Answer» Correct Answer - B In iodometry, `H_(2)Cr_(2)O_(7)` liberates `I_(2)` from iodides (Nal or Kl) which is titrated with `Na_(2)S_(2)O_(3)` solution `K_(2)Cr_(2)O_(7) + I^(-) + H^(+) rarr Cr^(3+) + I_(2)` Here, one mole of `K_(2)Cr_(2)O_(7)` accepts 6 mole of electrons. `:.` Equivalent weight `= ("Molecular weight")/(6)` |
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| 927. |
Convert the following in kilogram. (i) `0.91xx10^(-27)g` (Mass of electron) (ii) `3.34xx10^(-24)g` (Mass of hydrogen molecule) |
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Answer» Correct Answer - (i) `9.1xx10^(-3)kg` (ii) `3.34xx10^(-27)kg` |
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| 928. |
How many moles electron weigh one kilogramA. `6.023xx10^(23)`B. `1/9.108xx10^(31)`C. `6.023/9.108xx10^(54)`D. `1/(9.108xx6.023) xx10^(8)` |
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Answer» Correct Answer - D 1 mole of electron weight `9.108 xx 10^(-31) xx 6.023 xx 10^(23)kg` So number of moles of electron in 1 kg is `(1)/(9.108 xx 10^(-31) xx 6.023 xx 10^(23)) = (1)/(9.108xx 6.023) xx10^(8)` |
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| 929. |
Internatinal prototype of the kilogram is made up ofA. paladium-osmiumB. platinum-iridiumC. potassium-caseiumD. plutonium-indium |
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Answer» Correct Answer - B Pt-Ir-block |
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| 930. |
Express `(6.6xx10^(9))/(5.5xx10^(7))` in scientific notation. |
| Answer» Correct Answer - `1.2xx10^(2)` | |
| 931. |
Express the following in scientific notation. (1) 175000 (2) 0.17 |
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Answer» (1) `1.75xx10^(5)` (2) `1.7xx10^(-1)` |
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| 932. |
Express `(6.6xx10^(4))+(0.55xx10^(5))` in scientific notation. |
| Answer» Correct Answer - `12.1xx10^(4)=1.21xx10^(5)` | |
| 933. |
Express the following mathematical operations in scientific notation. (1) `(7.7xx10^(9))/(6.6xx10^(5))` (2) `(7.7xx10^(-7))/(6.6xx10^(-5))` |
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Answer» (1) `(7.7)/(6.6)xx10^(9-5)=1.16xx10^(4)` (2) `(7.7)/(6.6)xx10^(-7-5)=1.16xx10^(-12)` |
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| 934. |
Express the following mathematical operations in scientific notation. (1) `(7.7xx10^(4))+(0.77xx10^(5))` (2) `(8.7xx10^(4))-(0.77xx10^(5))` |
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Answer» (1) `(7.7xx10^(4))+(7.7xx10^(4))=15.4xx10^(4)` (2) `(8.7xx10^(4))-(7.7xx10^(4))=1xx10^(4)` |
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| 935. |
Express the following mathematical operations in scientific notation. (1) `(6.6xx10^(5))xx(7.7xx10^(9))` (2) `(6.6xx10^(5))xx(7.7xx10^(-7))` |
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Answer» (1) `6.6xx7.7xx10^(5+9)=50.82xx10^(14)` (2) `6.6xx7.7xx10^(5+(-7))=50.82xx10^(-2)` |
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| 936. |
Few figures are expressed in scientific notation. Mark the incorrect one.A. `234000=2.34xx10^5`B. `8008=8xx10^3`C. `0.0048=4.8xx10^(-3)`D. `500.0=5.00xx10^2` |
| Answer» Correct Answer - 8008=`8.008xx10^3` | |
| 937. |
Express `(2.2xx10^(4))xx(3.3xx10^(5))` in scientific notation. |
| Answer» Correct Answer - `7.26xx10^(9)` | |
| 938. |
Assertion: Scientific notation for the number 100 is expressed as `1xx10^2` Reason : The number `1xx10^2` has two significant figures.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» The terminal zeros are not significant if there is no decimal point. Hence, the number `1xx10^2` has only one significant figure. | |
| 939. |
Numbers are expressed in scientific notation. In this notation, every number is written as `N xx 10^(n)`, where `N = a` number with a single non-zero digit to the left of the decimal point `n = an` interger How many different scientific notations may be written for the result `10500g`A. FourB. threeC. FiveD. Two |
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Answer» Correct Answer - B `1.05xx10^(4), 1.050xx10^(4)`, and `1.0500xx10^(4)` |
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| 940. |
The amount of water that should be added to 500 ml of 0.5 N solution of NaOH to give a concentration of 10 mg per ml isA. (a)100B. (b)200C. (c )250D. (d)500 |
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Answer» Correct Answer - D `N_(1)=0.5 N rarr 10 g per ml` `N_(2)=(10xx10^(-3)g)/(40xx1)xx1000=0.25 N` `V_(1)=500 ml, V_(2)=?` `N_(1)V_(1)=N_(2)V_(2), 0.5xx500=0.25xxV_(2)` `V_(2)=1000 ml` final volume water added`=1000-500=500 mL`. |
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| 941. |
Which of the following is correct for luminous intensity?A. It is amount of light emitted per second in unit solid angle by a point source direction.B. The `SI` unit of luminous intensity is candelaC. The term "luminous intensity" is restricted to point sources only.D. All of these |
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Answer» Correct Answer - D Candela is one of the fundamental `SI` units. |
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| 942. |
Write the SI unit of luminous intensity and the amount of substance. |
| Answer» The SI unit of luminous intensity is candeia and that of amount of substance is mole. | |
| 943. |
12 g of Mg (at. Mass 24) will react completely with acid to giveA. One mole of `H_(2)`B. `1//2` mole of `H_(2)`C. `2//3` mole of `O_(2)`D. Both `1//2` mol of `H_(2)` and `1//2` mol of `O_(2)` |
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Answer» Correct Answer - B `Mg^(+2) equiv H_(2)` `n=(12 gm)/(24 gm)=1/2` mole of `H_(2)` |
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| 944. |
12 g of Mg (at. Mass 24) will react completely with acid to giveA. (a)One mole of `H_(2)`B. (b)`1//2` mole of `H_(2)`C. (c )`2//3` mole of `O_(2)`D. (d)Both `1//2` mol of` H_(2)` and `1//2`mol of` O_(2)` |
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Answer» Correct Answer - B `Mg^(+2)=H_(2)` `n=(12g)/(24g)=1/2` mole of `H_(2)` |
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| 945. |
Nitric acid `(HNO_(3))` is `1.6% H, 22.2 %N`, and `76.2% O` by mass. All pure samples of `HNO_(3)` have this compostion according to theA. law of mutliple proportionsB. law of reciprocal proportionsC. law of definite propotionsD. law of conservation of mass |
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Answer» Correct Answer - C According to the law of define proptions, all pure samples of a given compond always consist of the same elements combined in the same proportions by mass. |
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| 946. |
Which pair of species has the same percentage compostion?A. `C_(6) H_(12)O_(6)` and `C_(12) H_(22) O_(11)`B. `C_(6) H_(12)O_(6)` and `CH_(3) COOH`C. `C_(2) H_(5) OH` and `CH_(3) COOH`D. `C_(12) H_(22) O_(11)` and `HCOOH_(3)` |
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Answer» Correct Answer - B Compounds with different molecular formual will have the same percentage composition if they the same emprical formula. Both `C_(0) H_(12) O_(6)` and `CH_(3) CO_(2)H` have the same `EF`, i.e.,n `CH_(2)O`. |
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| 947. |
Amount of oxalic acid present in a solution can be determined by its titration with `KMnO_(4)` solution in the presence of `H_(2)SO_(4)`. The titration gives unsatisfactory result when carried out in the presence of HCl because HCl:A. (a)gets oxidized by oxalic acid to chlorineB. (b)Furnishes `H^(+)` ions in addition to those from oxalic acidC. (c )Reduces permanganate to `Mn^(2+)`D. (d)Oxidises oxalic acid to carbon dioxide and water |
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Answer» Correct Answer - C Titration of oxalic acid by `KMnO_(4)` in the presence of `HCl` gives unsatisfactory result because HCl is a better reducing agent than oxalic acid and `HCl` reduces preferably `MnO_(4)^(-) to Mn^(2+)` |
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| 948. |
How many moles of lead (II) chloride will be formed from a reaction between `6.5g` of `PbOP` and `3.2g` of `HCl` ?A. `0.333`B. `0.011`C. `0.044`D. `0.029` |
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Answer» Correct Answer - D According to balanced equation, `{:(PbO+,2HCl_((aq.))rarr,PbCl_(2)+,H_(2)O(l)),("1 mol","2 mol","1 mol","2 mol"):}` Calculating moles of reactants : `n_(PbO) = (6.5g)/(22.3g mol^(-1)) = 0.029 mol` `n_(NCl) = (3.2g)/(36.5g mol^(-1)) = 0.87 mol` Calculating moles of product formed from reactions: `0.029 mol PbO xx (1 mol PbCl_(2)) = 0.029 mol PbCl_(2)` `0.087 mol HCl xx (1 mol PbCl_(2))/(2 mol HCl) = 0.44 mol PbCl_(2)` Thus, `PvbO` is the limiting reactant and `0.029 moln PbCl_(2)` will be formed. |
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| 949. |
The moles of product are always determined by the starting moles ofA. excess reactantB. more reactive reactantC. limiting reactantD. less reactive reactant |
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Answer» Correct Answer - C Limiting recatant is entirely consumed when a reaction goes to completion. |
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| 950. |
An aqueous solution that is `3.5%` sodium chloride by mass can be prepared by dissolving `3.5g` of `NaCl` in `.........g` of `H_(2) O`A. `100`B. `103.5`C. `96.5`D. less than `100` |
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Answer» Correct Answer - C `3.5% NaCl` by mass contains `3.5g NaCl` in `100.0g` of solution. Thus, Mass of `H_(2) O` = (Mass of solution) - (mass of solute) `= (100.0 - 3.5)g = 96.5g` |
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