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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.
| 751. |
What is the mass of the precipitate formed when 50 mL of 16.9% solution of `AgNO_(3)` is mixed with 50 mL of 5.8% NaCl solution?A. 28 gB. 3.5 gC. 7 gD. 14 g |
| Answer» Correct Answer - C | |
| 752. |
Oxidation number of iodine varies fromA. `-1` to `+1`B. `-1` to `+7`C. `+3` to `+5`D. `-1` to `+5` |
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Answer» Correct Answer - B Iodine shows -1 to +7 oxidation state. |
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| 753. |
The number of moles of hydrogen that can be added to 1 mole of an oil is the highest inA. Linseed oilB. Groundanut oilC. Sunflower seed oilD. Mustard oil |
| Answer» Correct Answer - A | |
| 754. |
Which of the following has the highest mass?A. 20 g of sulphurB. 4 mol of carbon dioxideC. `12xx10^(24)` atoms of hydrogenD. `11*2` L of helium at N.T.P. |
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Answer» Correct Answer - B (B) Molar mass of `CO_(2) = 44 g mol^(-1)` `:.` 4 mol of `CO_(2) = 44xx4=176 g` (C) `6*02xx10^(23)` atoms of hydrogen has mass `=1*008 g` `12xx10^(24)` atoms of hydrogen has mass `=20*1g` (D) `22*4` L of helium at N.T.P. has mass = 4g 11*2 L of helium at N.T.P. has mass =2 g Thus 4 moles of `CO_(2)` has maximum mass. |
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| 755. |
Assuming that the density of water to be `1g//cm^(3)`, calculate the volume occupied by one molecule of water.A. `2.989 xx 10^(-23)`mLB. `6.023 xx 10^(23)` `cm^(3)`C. `0.288 xx 10^(-3)`D. `1.66 xx 10^9-2)` |
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Answer» a) `therefore` Density of water `=1 g//cm^(3)` `therefore |
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| 756. |
If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximatelyA. `18 cm^(3)`B. `22400 cm^(3)`C. `6*02xx10^(-23) cm^(3)`D. `3*0xx10^(-23) cm^(3)` |
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Answer» Correct Answer - C Mass of 1 molecule of water `=("GMM")/(N_(0))=(18)/(6*02xx10^(23))g` `:.` Volume of 1 molecule of water `= ("Mass")/("Density")` `=(18xx10^(-23))/(6*02xx1)=3*0xx10^(-23) mL`. |
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| 757. |
Chlorine gas is prepared by reaction of `H_2SO_4` with `MnO_2` and NaCl. What volume of `Cl_2` will be produced at STP if 50 g of NaCl is taken in the reaction ?A. 1.915 LB. 22.4 LC. 11.2 LD. 9.57 L |
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Answer» `2NaCl+MnO_2 + 3H_2SO_4 to 2NaHSO_4 + MnSO_4 + Cl_2 + 2H_2O` `{:("2 moles","1 mole"),("(2 x 58.5 = 117 g )", "22.4 L (STP)"):}` 117g of NaCl `-= 22.4L " of " Cl_2` 50g of NaCl`-=(22.4)/(117)xx50=9.57 L " of " Cl_2` at STP |
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| 758. |
Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide. How much volume of `CO_2` will be obtained by thermal decomposition of 50g `CaCO_3` ?A. 1LB. 11.2 LC. 44 LD. 22.4 L |
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Answer» `underset(100g)(CaCO_3) to CaO + underset ("1 mole " = 22.4L)(CO_2)` 100g of `CaCO_3` at STP gives 22.4 L of `CO_2` 50g of `CaCO_3` will produce `(22.4)/(100)xx50=11.2L " of " CO_2` |
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| 759. |
If isotopic distribution of C-12 and C-14 is 98% and 2% respectively, then the number of C-14 atoms in 12 g of carbon isA. `1*032xx10^(22)`B. `3*01xx10^(22)`C. `5*88xx10^(23)`D. `6*02xx10^(23)` |
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Answer» Correct Answer - A In 12 g of carbon, the amount of C-14 `=12xx2//100=0*24g` `:.` C-14 atoms in `0.24g = (0*24xx6*02xx10^(23))/(14)` `= 1*03xx10^(22)` atoms. |
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| 760. |
Earlier the concept of equivalent weight was very common and the concentrations of the solutions were expressed in terms of normolities. The convenience was that the substances reacted in the ratio of their gram equivalents. So there was no need for writing the balanced equations to determine the amounts of the substances reacted. However, determination of equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions. For example, `KMnO_(4)` has different equivalent weight in the basic medium than in teh acidic medium. Hence, now-a-days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality, molarity, mole fractions etc. are also used The molality of the above solution will be nearlyA. `15.3 m`B. `16.3 m`C. `17.3 m`D. `18.3 m` |
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Answer» Correct Answer - D Mass of the solvent `=100-40=60g` `=0.060kg` Moalrity `= ("Moles of solute")/("Solvent in kg")=(1.096mol)/(0.060kg)` `=18.3` |
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| 761. |
Earlier the concept of equivalent weight was very common and the concentrations of the solutions were expressed in terms of normolities. The convenience was that the substances reacted in the ratio of their gram equivalents. So there was no need for writing the balanced equations to determine the amounts of the substances reacted. However, determination of equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions. For example, `KMnO_(4)` has different equivalent weight in the basic medium than in teh acidic medium. Hence, now-a-days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality, molarity, mole fractions etc. are also used The mole fraction of hydrochloric acid in the solution will beA. `0.25`B. `0.30`C. `0.35`D. `0.40` |
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Answer» Correct Answer - A Moles of the solvent (water) `=(60)/(18)=3.33` Mole fraction of HCl in solution `=(1.096)/(1.096+3.33)=0.25` |
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| 762. |
Earlier the concept of equivalent weight was very common and the concentrations of the solutions were expressed in terms of normolities. The convenience was that the substances reacted in the ratio of their gram equivalents. So there was no need for writing the balanced equations to determine the amounts of the substances reacted. However, determination of equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions. For example, `KMnO_(4)` has different equivalent weight in the basic medium than in teh acidic medium. Hence, now-a-days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality, molarity, mole fractions etc. are also used The equivalent mass of CuA. will be the same in CuO and `Cu_(2)O`B. will be double in `Cu_(2)O` than in `CuO`C. will be double in `CuO` than in `Cu_(2)O`D. depends on whether copper is pure or impure |
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Answer» Correct Answer - B Eq. mass `=("At. Mass")/("Valency")` Eq. mass of `Cu(I)` in `Cu_(2)O=("At.Mass")/(1)` Eq. mass of Cu(II) in CuO = `("At. Mass")/(2)` Thus, eq. mass of Cu(I) in `Cu_(2)O = 2xx"Eq. mass of Cu(II) in CuO"` |
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| 763. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) The total charge in couloms required to complete the electrolysisA. `24125`B. `48250`C. `96500`D. `19300` |
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Answer» Correct Answer - D `2Cl^(-)to2Cl_(2)+2e^(-)` Total charge needed for completion of the electrolyses `=2xxF=2xx96500` coulomb `=193000` coulomb |
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| 764. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) The total number of moles of chlorine gas evolved isA. `0.5`B. `1.0`C. `2.0`D. `3.0` |
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Answer» Correct Answer - B 4.0 M NaCl, 500 mL (0.5L) Moles of NaCl `=0.5xx4=2` 2 moles of NaCl = 2 moles of `Cl^(-)` `underset("2 moles")(2Cl^(-))tounderset("1 mole")(Cl_(2))` `:.` No. of moles of `Cl_(2)` evolved = 1 mole |
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| 765. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) If cathode is a Hg electrode, the maximum weight(g) of amalgam formed from the solution isA. `200`B. `225`C. `400`D. `446` |
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Answer» Correct Answer - D `underset("2 moles")(Na+Hg)tounderset("2 moles")(Na-Hg)` `=23xx2+2xx200` `=446g` `:.` Maximum mass of amalgam formed = 446 g |
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| 766. |
If avogado number `N_(A)`, is changed from `6.022xx10^(23) mol^(-1)` to `6.022xx10^(20) mol^(-1)` this would changeA. The definition of mass in units of gramsB. The mass of one mole of carbonC. the ratio of chemical species to each other in a balanced equationD. The ratio of elements to each other in a compound |
| Answer» Correct Answer - B | |
| 767. |
Concentrated aqueous solution of sulphuric acid is `98 %` by mass and has density of `1.80 "g mL"^(-1)`. What is the volume of acid required to make one liter `0.1 M H_(2)SO_(4)` solution ?A. (a)`16.65 mL`B. (b)`22.20 mL`C. (c )`5.55 mL`D. (d)`11.10 mL` |
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Answer» Correct Answer - c Molarity of `H_(2)SO_(4)` solution `=(98xx1000)/(98xx100)xx1.802=18.02` Suppose V ml of this `H_(2)SO_(4)` is used to prepare 1 litre of 0.1 `M H_(2)SO_(4)`. Using formula `M_(1)V_(1)=M_(2)V_(2)` `:. Vxx18.02=1000xx0.1` or `V=(1000xx0.1)/18.02=5.55 ml` |
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| 768. |
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 properties of its constituent elements.D. The ratio of atoms of different elements in a compound is fixed |
| Answer» c) physical properties of a compound are different from those of elements. | |
| 769. |
The unit `J Pa^(-1)` is equivalent toA. (a)`m^(3) `B. (b)` cm^(3)`C. (c )` dm^(3)`D. (d)None of these |
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Answer» Correct Answer - A `Jpa^(-1)`, Unit of work is Joule and unit of pressure is Pascal. Dimension of Joule, i.e., work `=FxxL=MLT^(2)xxL=[M^(2)T^(-2)]` `1/(Pa)=1/Pressure=1/(F/A)=(1xxA)/F=[MLT^(-1)]` So, `Jpa^(-1)=[M^(2)T^(2)]=[L^(2)xxL]=[L^(3)]` |
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| 770. |
The unit `J Pa^(-1)` is equivalent toA. `m^(3)`B. `cm^(3)`C. `dm^(3)`D. None of these |
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Answer» Correct Answer - A `Jpa^(-1)`, unit of work is Joule and unit of pressure is Pascal. Dimension of Joule i.e. work `=FxxL=MLT^(-1) xxL` `=[ML^(2) T^(-2)]` `L^(2)/(MLT^(-2))=M^(-1) LT^(2)` `ML^(2)T^(-2)xxM^(-1)LT^(2)=L^(3)` |
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| 771. |
Which of the following statements is trud about the science of atoms are molecules?A. We can see, weigh and perceive the atoms and molecules through naked eyes.B. It is possible to count the number of atoms and molecules in a given mass of matter through naked eyes and manuallyC. We can establish a quantitative relationship between the mass and number of these particles ( atoms and molecules)D. physical properties of matter can be qualitatively described using numerical values with suitable units |
| Answer» Correct Answer - c) | |
| 772. |
Number of significant figures in 78.000 g, 0.0206 g and 3.002 g respectively areA. 3,4 and 5B. 2,5 and 4C. 3,3 and 4D. 5,3 and 4 |
| Answer» Correct Answer - d) | |
| 773. |
`81.4 g` sample of enthyl ethyl alcohol contains `0.002 g` of water. The amount of pure ethyl alcohol to the proper number of significant figures isA. 81.4B. 71.40 gC. 91.4 gD. 81 g |
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Answer» Correct Answer - A Pure ethyl alcohol `=81.4-0.002=81.398`. |
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| 774. |
The oxide which cannot act as reducing agent isA. `SO_(2)`B. `NO_(2)`C. `CO_(2)`D. `ClO_(2)` |
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Answer» Correct Answer - C `CO_(2)` is an acidic oxide. |
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| 775. |
Which substance is serving as a reducing agent in the following reaction? `14H^(+)+Cr_(2)O_(7)^(2-)+3Ni rarr 2Cr^(3+)+7H_(2)O+3Ni^(2+)`A. `H_(2)O`B. NiC. `H^(+)`D. `Cr_(2)O_(7)^(2-)` |
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Answer» Correct Answer - B The oxidation number of Ni changes from 0 to +2. |
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| 776. |
The ion(s) that act/s as oxidizing agent in solution is/areA. `Tl^(+)` and `Al^(3+)`B. `B^(3+)` and `Al^(3+)`C. `Tl^(3+)` onlyD. `B^(3+)` only |
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Answer» Correct Answer - C `Tl^(+)` ions are more stable than `Tl^(3+)` ions and thus, `Tl^(3+)` ions change to `Tl^(+)` ion thereby acting as oxidising agent. `underset("(Less stable oxidising agent)")(Tl^(3)" Compounds ")+2e^(-) rarr Tl^(+)underset(("(More stable"),("reducing agent)"))("Compounds")` |
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| 777. |
The simplest formula of a compound containing 50% of element X (atomic mass 10) and 50% of element Y (atomic mass 20) isA. `XY`B. `X_(2)Y`C. `XY_(3)`D. `X_(2)Y_(3)` |
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Answer» Correct Answer - B `|{:("Element",%(a),"At.wt"(b),a//b,"Rate"),(X,50,10,5,2),(Y,50,20,2.5,1):}|` Simple formula `= X_(2)Y` |
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| 778. |
The compound which could not act both as oxidising and reducing agent isA. `SO_(2)`B. `MnO_(2)`C. `Al_(2)O_(3)`D. `CrO` |
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Answer» Correct Answer - C `Al_(2)O_(3)` could not act as oxidizing and reducing agent |
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| 779. |
Which of the following is the most powerful oxidizing agent?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
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Answer» Correct Answer - A Fluorine is a most powerful oxidizing agent because it has `E^(@) = +2.5` volt |
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| 780. |
One part of element A reacts with two parts of another element B. 6 parts of element C reacts with 4 parts of element B. If A and C combine together, the ratio of their weights be governed byA. Law of conservation of massB. law of reciprocal proportionsC. law of definite proportionsD. law of multiple proportions |
| Answer» b) Ratio will be goverened by the law of reciprocal proportions. | |
| 781. |
The law of multiple proportion was proposed byA. LavoisierB. DaltonC. ProustD. Gay-Lussac |
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Answer» Correct Answer - B Law of multiple proportion was proposed by Dalton and verified by Berzelius |
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| 782. |
The law of multiple proportion is lillustrated by the pair of compoundsA. sodium chlordie and sodium bromideB. water and heavy waterC. sulphur dioxide and sulphur trioxideD. magnesium hydroxide and magnesium oxide |
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Answer» Correct Answer - C The law of multipw proportion is illustrated by the pair of coompounds : `SO_(2) and SO_(3).` |
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| 783. |
One part of an element A combines with two parts of another element B,6 parts of element C combines with 4 parts of B. if A and C combine together the ratio of their weights, will be governed byA. law of definite proportionsB. law of multiple proportionsC. law of reciprocal proportionsD. law of conservations of mass |
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Answer» Correct Answer - C The ratio of given masses will be governed by law of reciprocal proportions. |
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| 784. |
A mixture of `NaCl` and `Na_(2)CO_(3)` is given. On heating `12 g` of the mixture with dilute `HCl, 2.24 g` of `CO_(2)` is removed. Calculate the amounts of each in the mixture.A. `6.6` gB. `5.8` gC. `6.8` gD. `7.2` g |
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Answer» Correct Answer - A `CO_(2)` is from `Na_(2) CO_(3)` only `underset(underset((?))(106g))(Na_(2)CO_(3)) + 2HCl rarr2 NaCl + underset(underset(2,24 g)(44 g)) (CO_(2))+ H_(2) O ` 44 g `CO_(2)` is from = 106 g `Na_(2) CO_(3)` `therefore 2.24 g CO_(2) " is from " = 106/44 xx 2.24 " g " Na_(2) CO_(3) = 5.4g` Thus, `NaCl = 12 - 5.4 = 6.6` g |
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| 785. |
1 g of hydrogen is found to combine with 80g of bromine and 1g of calcium combines with 4 g of bromine. Equivalent weight of calcium isA. 16B. 20C. 40D. 80 |
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Answer» 1g `H_(2)` combines with `Br_(2)` = 80g Equivalent weight of `Br_(2)` = 80 4g of `Br_(2)` combine with = 1g Ca `therefore` 80 g of `Br_(2)` will combine with (`=1 xx 80)/4 = 20g Ca |
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| 786. |
The volume occupied by 1 mole of H atoms at NTP isA. `22.4` LB. `11.2` LC. `40.2` LD. None ot these |
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Answer» Correct Answer - D The volume occupied by 1 mole of H atoms at NTP is `11.2 ` L. |
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| 787. |
The weight of oxygen that will react with 1 g of calcium isA. `0.2` gB. `0.6` gC. `0.4` g CD. `0.8` g |
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Answer» Correct Answer - C Atomic mass of calcium = 40 g `underset(underset(underset(40 g)(1xx 40 g))(1" mol"))(Ca(s)) + underset(underset(underset(16 g)(8xx2))(1" mol"))(O_(2)(g)) rarr underset(underset(underset(underset(112 g)(2(56xx2)))(2(40+16)))(2" mol" )) (2CaO)` 40 g of Ca reacts with 19 g of oxygen to form 112 g of CaO. 1 g of Ca reacts with `16/40g = 0.4` g of oxygen. |
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| 788. |
By heating `10 g CaCO_(3), 5.6 g CaO` is formed. What is the weight of `CO_(2)` obtained in this reactionA. `2.4` gB. `5.6` gC. `4.4` gD. `3.6` g |
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Answer» Correct Answer - C `underset(underset(=100g)((40+1216xx3)))(CaCO_(3)) rarr underset(underset(=56 g)(40+16))(CaO) +underset(underset(=44g)((12+16 xx 2))) (CO_(2))` 100 g of `CaCO_(3) rArr` 56 g of CaO 10 g of `Ca CO^(3) rArr 5.6` g of CaO 100 g of `CaCO_(3) rArr` 44 g of `CO_(2)` 10 g of `CaCO_(3) rArr 4.4` g of ` CO_(2)` |
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| 789. |
500 mL of 0.250 M `Na_(2)SO_(4)` solution is treated with 15.00 g of `BaCl_(2)`. Moles of `BaSO_(4)` formed areA. `0.72`B. `0.072`C. `0.168`D. `0.0168` |
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Answer» Correct Answer - B `Na_(2)SO_(4)+BaCl_(2)toBaSO_(4)+2NaCl` Molar mass of `BaCl_(2) = 137.2+2xx35.5` `= 208.4 g mol^(-1)` Moles of `BaCl_(2) = (15)/(208.4g mol^(-1)) = 0.072 mol` Moles of `Na_(2)SO_(4) = ((500)/(1000)L)xx(0.250 mol L^(-1))` `= 0.125 mol` Here `BaCl_(2)` is the limiting reactant `:.` Moles of `BaSO_(4)` formed = Moles of `BaCl_(2)` `= 0.072 mol` |
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| 790. |
Assertion (A): Sodium perxenate `(Na_(4)XeO_(6))` reacts with `NaF` in acidic medium to give `XeO_(3)` and `F_(2)` Reason (R ): `XeO_(6)^(4-)` is a stronger oxidant than `F_(2)`.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 5B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 5C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A `overset(+8)(XeO_(6)^(4-))+overset(-1)(2F^(ɵ))+6H^(o+) rarr overset(+6)(Xe)O_(3)+overset(0)F_(2)+3H_(2)O` The oxidation number of Xe decreases from +8 to +6 in `XeO_(3)` while that of F increase from -1 to zero. Therefore, `XeO_(6)^(4-)` is reduced while `F^(ɵ)` is oxidised. Hence, `XeO_(6)^(4-)` is a stronger ocidant than `F_(2)` |
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| 791. |
In the reaction `2CuSO_(4)+4KI rarr 2Cu_(2)I_(2)+I_(2)+2K_(2)SO_(4)` the equivalent weight of `CuSO_(4)` will be:A. (a)`79.75`B. (b)`159.5`C. (c )`329`D. (d)None of these |
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Answer» Correct Answer - B `2CuSO_(4)+4KI rarr Cu_(2)I_(2)+I_(2)+2K_(2)SO_(4)`. `Cu^(2+)+1 e^(-) rarr Cu^(+)`. `E_(Cu)=? V.F. =1`. `E_(CuSO_(4))=159.5/1=159.5` |
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| 792. |
Chemical reactions involve interaction of atoms and molecules. A large number of atoms and molecules (approximately `6.022 xx 10^(23)`) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such a large number conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry adn radiochemistry. The following examples illustrate a typical case involving chemical/electrochemical reaction which requires a clear understanding of mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of the solution is electrolysed. This lead to the evolution of chlorine gas at one of electrodes (atomis mass : Na = 23 , Hg = 200 , 1F = 96500 C) If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from the solution is :A. 200B. 225C. 400D. 446 |
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Answer» No. of moles of `Na^+=2` 2NaCl `-=2Na` `underset("moles" 2)(Na)+ underset(2)(Hg) to underset(2)(Na(Hg))` By electrolysis we can get a maximum of 2 moles of sodium which can combine with exactly 2 moles of mercury to give 2 moles of amalgam. ltbr. `therefore` Maximum weight of Na amalgam (assuming equimolar Na and Hg) = 46 + 400 = 446 g. |
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| 793. |
Chemical reactions involve interaction of atoms and molecules. A large number of atoms and molecules (approximately `6.022 xx 10^(23)`) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such a large number conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry adn radiochemistry. The following examples illustrate a typical case involving chemical/electrochemical reaction which requires a clear understanding of mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of the solution is electrolysed. This lead to the evolution of chlorine gas at one of electrodes (atomis mass : Na = 23 , Hg = 200 , 1F = 96500 C) The total number of moles of chlorine gas evolved is :A. 0.5B. 1C. 2D. 3 |
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Answer» 500 mL of 4.0 molar NaCl solution contains 2 moles of NaCl. The chlorine content of this sample will be evolved as chlorine gas. The number of moles of NaCl = number of moles of Cl = 2 moles `therefore` Number of moles of `Cl_2` gas evolved `=2/2=1mole (2NaCl to Cl_2)` |
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| 794. |
The number of molecules in 8.96L of a gas at `0^(@)C` and 1 atmosphere pressure is approximatelyA. `6.02xx10^(23)`B. `12.04xx10^(23)`C. `18.06xx10^(23)`D. `24.08xx10^(22)` |
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Answer» Correct Answer - D No. of molecules `=(8.96)/(22.4)xx6.02xx10^(23)=2.408xx10^(23)` `=24.08xx10^(22)` |
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| 795. |
Under S.T.P. 1 mol of `N_(2)` and 3 mol of `H_(2)` will form on complete reactionA. 4 moles of `NH_(3)`B. 89.6 L of `NH_(3)`C. 22.4 L of `NH_(3)`D. 44.8 L of `NH_(3)` |
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Answer» Correct Answer - D `underset(1mol)(N_(2))+underset(3mol)(3H_(2))tounderset(2mol)(2NH_(3))` 1 mol of `N_(2)` reacts with 3 moles of `H_(2)` to give 2 moles of `NH_(3)` which occupies a volume `= 2xx22.4 = 44.8 L` at S.T.P. |
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| 796. |
`100 mL` of phosphine `(PH_(3))` on hearing forms phosphorous `(P)` and hydrogen `(H_(2))`. The volume change in the reaction isA. an increase of 50 mLB. an increase of 110 mLC. an increase of 150 mLD. a decrease of 50 mL |
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Answer» Correct Answer - A `underset(100mL)underset(2mL)(2PH_(3))(g)to2P(s)+underset(150mL)underset(3mL)(3H_(2))(g)` `:.` 100 mL of `PH_(3)` gives `H_(2)=150mL` Increase in volume `=150-100=50mL` |
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| 797. |
100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume isA. 50 mL increaseB. 500 ml decreaseC. 900 mL decreaseD. nil |
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Answer» Correct Answer - A `underset(2vol)(2PH_(3))tounderset(3vol)(3H_(2))+2P` 2 vol of `PH_(3)` give 3 vol of `H_(2)` `:.` 100 mL of `PH_(3)` gives 150 mL of `H_(2)` `:.` Increases in volume = 150 mL - 100 mL = 50 mL |
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| 798. |
The decomposition of a certain mass of `CaCO_(3)` gave `11.2 dm^(3)` of a `CO_(2)` at S.T.P. The mass of KOH required to completely neutralize the gas isA. `56 g`B. `28g`C. `42g`D. `20g` |
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Answer» Correct Answer - A `underset(112g)underset(2(39+16+1))(2KOH)+underset("at S.T.P.")underset(22.4 L)(CO_(2))toK_(2)CO_(3)+H_(2)O` `22.4 dm^(3)` of `CO_(2)` at N.T.P. required KOH = 112 g `11.2 dm^(3)` of `CO_(2)` at S.T.P. will require KOH = 56 g |
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| 799. |
What mass of `CaCO_(3)` is required to react completely with 25 ml of `0.75 MHCI`?A. `0.94 g`B. `9.4 g`C. `0.094 g`D. `0.49 g` |
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Answer» Correct Answer - A `CaCO_(3) + 2HCl to CaCl_(2) + CO_(2) + H_(2)` 25 mL of 0.75 M HCl `= (25)/(1000)Lxx(0.75 mol L^(-1))` `= 0.01875 mol` Moles of `CaCO_(3)` required `= ("Moles of HCl")/(2)` `= (0.01875)/(2) = 9.375xx10^(-3) mol` Mass of `CaCO_(3)` required `= 9.375xx10^(-3) molxx100gmol^(-1)` `= 0.9375g = 0.94g` |
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| 800. |
A hydrocarbon contains `10.5 g` of carbon per gram of hydrogen. `1 L` of vapour of the hydrocarbon at `127^(@)C` and 1 atm pressure weighs `2.8 g`. Find the molecular formula of the hydrocarbon.A. `C_(6)H_(14)`B. `C_(5)H_(10)`C. `C_(6)H_(12)`D. `C_(7)H_(8)` |
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Answer» Correct Answer - D Using the general gas equation `: pV = nRT`, we have `1xx 1 = (2.8)/m xx 0.0821 xx 400 ` Hence, molar mmass `= 91.95` Atomic ratio `H = 1/1, C= 10.5/12 = 0.875 ` Lest whole number ratios: H = 8 and C = 7 (`0. 875` multiplied by 8 gives 7 as whole figure) Hence, empirical formula `= C_(7)H_(8)` Empirical formula weight = 92 Molar mass and empirical formula mass being the same, the molecular formula is `C_(7) H_(8) |
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