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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.
| 301. |
Which of the following is a homongenous mixture?A. MilkB. A mixture of two minerals such as galena (black) and quartz (white)C. Blue copper (II) sulphate solutionD. Foggy air |
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Answer» Correct Answer - C It is an aqueous solution of copper `(II)` sulphate. Milk and foggy air are colloidate solutions. |
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| 302. |
Assertion: As mole is the basic chemical unit, the concentration of the dissolved solute is usually specified in terms of number of moles of solute. Reason: The total number of molecules of reactants involved ina balanced chemical equation is known as molecularity of the reaction.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 the 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 The number of moles of a solute present in one litre of solution is known is as molarity (M). The total number of molecules of reactants present in a balenced chemical equation is known as molecularity. For example, `PCl_(5) rarr PCl_(3)+Cl_(2)` (Unimilecular) `2HCl rarr H_(2)+I_(2)` (Bimolecular) `:.` Molarity and molecularity are used in different sense. |
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| 303. |
Which of the following terms are unitless?A. MolalityB. MolarityC. Mole fractionD. Mass per cent |
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Answer» Both moel fraction and mass per cent are unitless as b oth are ratios of moles and mass respectively. Mole fraction `=("Nuumber of moles of solute")/("Number of moles of solution")=("moles")/("moles")` `=("Number of moles of solvent")/("Number of moles of Solution")=("moles")/("moles")` `"Mass per cent"=("Mass of soulte in gram")/("Mass of solution in gram")xx100` |
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| 304. |
Total number of atoms in 44 g `CO_(2)` isA. `6*02xx10^(23)`B. `6*02xx10^(24)`C. `1*806xx10^(22)`D. `18*06xx10^(22)`. |
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Answer» Correct Answer - C Total no. of moles of `CO_(2)` in 44 g `= (44)/(44) = 1` 1 mole `CO_(2)` = 1 mole C + 2 mole O atoms = 3 mole atom `= 3xx6*02xx10^(23) = 1*806xx10^(24)` |
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| 305. |
If we take 44 g of `CO_(2)` and 14 g of `N_(2)`. What will be the mole fraction of `CO_(2)` in the mixture?A. `1/5`B. `1/3`C. `2/3`D. `1/4` |
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Answer» Number of moles of `CO_(2)=(44)/(44)=1` Number of moles of `N_(2)=(14)/28)=0.5` `therefore` Mole of fraction of `CO_(2)=(1)/(1+0.5)=(1)/(1.5)or= (2)/(3)` |
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| 306. |
The mass of a molecule of water isA. (a)`3xx10^(-26) kg`B. (b)`3xx10^(-25) kg`C. (c )`1.5xx10^(-26) kg`D. (d)`2.5xx10^(-26) kg` |
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Answer» Correct Answer - A `6xx10^(23)` molecules has mass`=18 g` 1 molecules has mass`=18/(6xx10^(23))` `=3xx10^(-23) g` `=3xx10^(-26) kg` |
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| 307. |
In the multiplication of2.8 and 4.5039, the correct number of significiant figures in the reported answer will beA. fiveB. twoC. sixD. seven |
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Answer» Correct Answer - B `(2.8)xx(4.5039) = 12.61092`, round off to 13. |
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| 308. |
In the division of 6.85 by 112.04, the correct number of significiant figures in the reported answer will beA. fiveB. nineC. eightD. three |
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Answer» Correct Answer - D `(6.85)div (112.04) = 0.0611388789`, round off to `0.0611`. |
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| 309. |
Assertion : One mole of `SO_(2)` contains double the number of molecules present in one mole of `O_(2)` Reason : Molecular weight of `SO_(2)` is double to that of `O_(2)`.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 assertion is false but reason is true. |
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Answer» Correct Answer - D One mole of any substance corresponding to `6.023xx10^(23)` entities is respective of its weight. Molecular weight of `SO_(2)=32+2xx16=64 gm`. Molecular weight of `O_(2)=16xx2=32 gm`. `:.` Molecular weight of `SO_(2)` is double to that of `O_(2)`. |
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| 310. |
A gaseous mixture contain `CH_(4)` and `C_(2)H_(6)` in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP isA. (a)4.6 gB. (b)2.3 gC. (c )1.6 gD. (d)23 g |
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Answer» Correct Answer - B Equimolecular proportion means both gases occupied equal volume`=2.24/2=1.12L` For `CH_(4):` `22.4L CH_(4)` has mass`=16 g` `1.12L CH_(4)` has mass`==16/22.4xx1.12=0.8 g` For `C_(2)H_(6)` `22.4L C_(2)H_(6)` has mass`=30 g` `1.12L C_(2)H_(6)` has mass `30/22.4xx1.12=3.0/2 g=1.5 g` Total mass`=1.5g +0.8g=2.3 g`. |
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| 311. |
Assertion : 22.4 L of `N_(2)` at NTP and 5.6 L `O_(2)` at NTP contain equal number of molecules. Reason : Under similar conditions of temperature and pressure all gases contain equal number of molecules.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 assertion is false but reason is true. |
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Answer» Correct Answer - D Molar volume (at NTP) `=22.4 L` Now 22.4 L of `N_(2)=` volume occupied by one mole of `N_(2)=28` gm `=6.023xx10^(23)` molecules. Similarly, `O_(2)=2xx16=32 gm`, `32 gm=6.023xx10^(23)` molecules `=22.4 L` `:. 22.4 L =6.023xx10^(23) or 5.6 L =(6.023xx10^(23)xx5.6)/22.4` `=1/4xx6.023xx10^(23)` |
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| 312. |
The flask A and B of equal size contain 2 g of `H_(2)` and 2 g of `N_(2)` respectively at the same temperature. The number of molecules in flask A isA. same as those in flask BB. less than those in flask BC. greater then those in flask BD. exactly half than those in flask B |
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Answer» Correct Answer - C Moles of `H_(2) = (2)/(2) = 1` Moles of `N_(2) = (2)/(28) = (1)/(14) = 0.0714` |
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| 313. |
Barn is used to measure the cross-sectional area of the mucles of an atom. IT is equal toA. `10^(-28) m^(2)`B. `10^(-25) m^(2)`C. `10^(-31) m^(2)`D. `10^(-17) m^(2)` |
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Answer» Correct Answer - C Barn is the unit of area for measurring the cross-section of nuclei. 1 barn equals `10^(-24) cm^(2)` or `10^(-28) m^(2)`. |
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| 314. |
Correct order of prefixes isA. pico `lt` nano `lt` micro `lt` milliB. nano `lt` pico `lt` micro `lt` milliC. micro `lt` pico `lt` milli `lt` nanoD. milli `lt` micro `lt` pico `lt` nano |
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Answer» Correct Answer - A `10^(-12) ("pico") gt 10^(-9) ("nano") gt 10^(-6) ("micro") gt 10^(-3) ("milli")` |
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| 315. |
1 mol of `CH_(4)` containsA. 4 g atoms of hydrogenB. `3.0` g atoms of carbonC. `6.02 xx 10^(23)` atoms of hydrogfenD. `1.81 xx10^(23)` molecules of `CH_(4)` |
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Answer» Correct Answer - A 1 mol of `CH_(4) = 16` g of `CH_(4)`[molar mass] Thus, 1 mol of `CH_(4) ` contains 4 g atoms of hydrogen. |
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| 316. |
`0.1 mol` of a carbonhydrate with empirical formula `CH_(2)O` contains `1 g` of hydrogen. What is its molecular formula?A. `C_(5)H_(10)O_(5)`B. `C_(6)H_(12)O_(6)`C. `C_(4)H_(8)O_(4)`D. `C_(3)H_(6)O_(3)` |
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Answer» Correct Answer - A 0.1 mol of cabohydrate with empirical formula `CH_(2)O` contains 1 g of hydrogen. `:.` 1 mol of carbohydrate with empirical formula In `CH_(2)O` atomic ratio of `C:H:O=1:2:1` with 10 g atoms of H, g atoms of carbon = 5 and g atoms of oxygen =5 `:.` The molecular formula of the carbohydrate is `C_(5)H_(10)O_(5)`. |
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| 317. |
Zinc sulphate contains `22.65%` of zinc and `43.9%` of water of crystallization. If the law of constant proportions is true, then the the weight of zinc required to produce `20 g` of the crystals will beA. (a)`45.3 g`B. (b)`4.53 g`C. (c )`0.453 g`D. (d)`453 g` |
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Answer» Correct Answer - B `100 g` of `ZnSO_(4)` crystals are obtained from`=22.65 g Zn` `1 g` of `ZnSO_(4)` crystals will be obtained from`=22.65/100 g Zn` `20 g` of `ZnSO_(4)` crystals obtained from`=22.65/100xx20=4.53 g` |
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| 318. |
Which of the following is not equal to one gram?A. `1000 mg`B. `1000000 mu g`C. `10000 dg`D. `100 cg` |
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Answer» Correct Answer - C `m (10^(-3)), mu (10^(-6)), d (10^(-1)), c (10^(-2))` |
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| 319. |
Which of the following is not correct regarding `SI` derived units?A. Pressure in pascal (pa)B. Density in kilogram per cubic meter `(kg m^(-3))`C. Electric charge in coulomb `(C)`D. Energy in electron volt `(eV)` |
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Answer» Correct Answer - A The `Sl` unit of energy is joule `(J)`. It is defined as the work when the point of application of a force of one newton is displaced through a distance of one meter in the direction of the force. |
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| 320. |
Determine the empirical formula `(EF)` of the oxide of chomium containing `68.4% Cr` by mass.A. `Cr O_(3)`B. `Cr_(2) O_(3)`C. `Cr O_(5)`D. `Cr_(3) O_(4)` |
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Answer» Correct Answer - B `n_(Cr) = (68.4g)/(52g mol^(-1)) = 1.32, n_(O) = (31.6g)/(16g mol^(-1)) = 1.98` `(n_(Cr))/(n_(O)) = (1.32)/(1.98) = (132)/(198) = (2xx66)/(3xx66) = (2)/(3)` `:. EF` is `Cr_(2) O_(3)`. |
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| 321. |
0.1 " mol of "`MnO_(4)^(ɵ)` (in acidic medium) can:A. Oxidies 0.5 mol of `Fe^(2+)`B. Oxidise 0.166 mol of `FeC_(2)O_(4)`C. Oxidise 0.25 mol of `C_(2)O_(4)^(-2)`D. Oxidise 0.6 mol of `Cr_(2)O_(7)^(-2)` |
| Answer» Correct Answer - A::B::C | |
| 322. |
The element exhibiting most stable +2 oxidation state among the following isA. `Ag`B. `Fe`C. `Sn`D. `Pb` |
| Answer» Correct Answer - D | |
| 323. |
Mixture of sand and sulphur may best be separated byA. Fractional crystallisation from aqueous solutionB. Magnetic methodC. Fractional distillationD. Dissolving in `CS_(2)` and filtering |
| Answer» Correct Answer - B | |
| 324. |
A mixture of sand and iodine can be separated byA. CrystallisationB. SublimationC. DistillationD. Fractional distillation |
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Answer» Correct Answer - B Iodine shows sublimation and hence volatalizes on heating the vapour condenses on colloing to give pure iodine. |
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| 325. |
Length conserversion: Angstrom `(Å)`, a unit of length `(1xx10^(-10)m)` is commonly used to describe the radii of atoms, which are often expressed in other unirs. Find the radius of a sillicon atoms `(1.17 Å)` in centimeters and nanometers. ltbgt Strategy: `Å rarr m rarr cm` `Å rarr m rarr nm` Use th e equations `1 Å = 1xx10^(-10) m, 1 cm = 1xx10^(-2) m`, and `1mm = 1xx10^(-9) m` to construct the unit factors that convert `1.17 Å` to the required units. |
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Answer» `? Cm = (1.17 Å) xx ((1xx10^(-10) m)/(1 Å)) xx ((1cm)/(1xx10^(-2) m))` Alternatively, `1.17 Å rarr 1.17 xx 10^(-10) m rarr 1.17 xx 10^(-19) xx100 cm` `? Nm = (1.17 Å) xx ((1xx10^(-10)m)/(1 Å)) xx ((1nm)/(1xx10^(-9) m))` Alternatively, `1.17 Å rarr 1.17 xx10^(-10) m rarr 1.17xx10^(-1)xx10^(-9) m` `rarr 1.17xx100 "nm"` |
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| 326. |
Which of the following statements about a compound is incorrect ?A. A molecule of a compound has atoms of different elements.B. A compound cannot be separated into its constituent elements by physical methods of separation.C. A compound retains the physical properties of its constituents elements.D. The ratio of atoms of different elements in a compound is fixed. |
| Answer» Physical and chemical properties of a compound are different from those of its constituent elements. | |
| 327. |
The empirical formula and molecular mass of a compound are `CH_(2)O` and 180 g respectively. What will be the molecular formula of the compound ?A. `C_9H_18O_9`B. `CH_2O`C. `C_6H_12O_6`D. `C_2H_4O_2` |
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Answer» `n=("Molecular mass")/("Empirical formula mass") = 180/30=6 ( therefore " Empirical formula mass of " CH_2O=30)` `therefore " Molecular formula " = 5xx CH_2O=C_6H_12O_6` |
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| 328. |
If the density of a solution is `3.12 "g mL"^(-1)`, the mass of 1.5 mL solution in significant figures isA. 4.7 gB. `4680xx10^(-3)` gC. 4.680 gD. 46.80 g |
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Answer» Density `=("Mass")/("Volume")` `therefore " Mass = Density" xx " Volume "` `3.12g mL^(-1) xx 1.5 mL = 4.68 g = 4.7g` |
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| 329. |
Match the following. |
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Answer» A. Number of moles of `CO_(2)` molecule `=("Weight in gram of" CO_(2))/("molecular weight of " CO_(2))=(88)/(44)=2` mol B. 1 mole of a substance `=N_(A)` molecules `=6.022xx10^(23)` molecuels =Avogadro number `=6.022xx10^(23)` molecuels of `H_(2)O=1` mol C. 22.4L of `O_(2)` at StP=1 mol D. Number of moles of 96g of `O_(2)=(96)/(32)"mol"=3"mol"` E. 1 mole of any gas=Avogadro number `=6.022xx10^(23)` molecuels |
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| 330. |
Which of the following is correct?A. `1 L = 1 m^(3)`B. `1 L = 1 dm^(3)`C. `10 L = 1 dm^(3)`D. `1 L = 10 dm^(3)` |
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Answer» Correct Answer - B 1L = `1000 cm^(3) = 10 cmxx10cmxx10cm` `=1dmxx1dmxx1dm=1dm^(3)` |
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| 331. |
Light travels with a speed of `3xx10^(8) m s^(-1)`. The distance travelled by light in 1 femto second isA. `0*03 mm`B. `0*003 mm`C. `3 mm`D. `0*0003 mm` |
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Answer» Correct Answer - D `S=Cxxt=(3xx10^(8)ms^(-1)xx(10^(-15)s)` `=3xx10^(-7)m` `=3xx10^(-7)xx10^(3)mm = 3xx10^(-4)mm` `=0*0003 mm` |
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| 332. |
A jug contains 2L of milk. Calcualte the volume of the milk in `m^(3)`A. `2 xx 10^(-2)m^(3)`B. `2 xx 10^(-1)`C. `2 xx 10^(-3)m^(3)`D. `2 xx 10^(-4)`m |
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Answer» c) `2 xx 10^(-3)m^(3)` `therefore` 1L = 1000 `cm^(3)` and 1m = 100 cm which gives `(1m)/(100cm)=1=(100cm)/(100cm)` To get `m^(3)` from the above unit factors, the first unit factor is taken and it is cubed `(1m)/(100 cm)^(3)` rArr (1m^(3)/(10^(6)cm^(3))=(1)^(3)=1` Now, 2L = 2 `xx 1000cm^(3)` The above is multiplied by unit factor `2 xx 1000 cm^(3) xx (1m^(3)/(10i^(6)cm^(3))= (2m^(3))/(10^(3))` = 2 xx 10^(-3)m^(3)` |
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| 333. |
Express the following numbers to four significant figures. (1) 6.608792 (2) 42.392800 |
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Answer» (1) As the fifth digit 7 is greater than 5, therefore the result will be expressed as 6.609. (2) It will be expressed as 42.39. The digit 2 is ddropped since it is less then 5 (the figure is not rounded off to the next number). |
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| 334. |
A compound having the empirical formula `(C_(3)H_(4)O)_(n)` has a molar mass of `170 +- 5`. The molecular formula of this compound isA. `C_(3)H_(4)O`B. `C_(6)H_(8)O_(2)`C. `C_(6)H_(12)O_(3)`D. `C_(9)H_(12)O_(3)` |
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Answer» Correct Answer - D `E.F. = C_(2)H_(4)O` E.F. mass `= 12xx3+4xx1+16 = 56` `n = (170+-5)/(56)~~3` `:. M.F. = (C_(3)H_(4)O)_(3) = C_(9)H_(12)O_(3)` |
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| 335. |
A gaseous hydrocarbon on complete combustion gives 3.38 g of `CO_(2)` and 0.690 g of `H_(2)O` and no other product. The empirical formula of the hydrocarbon isA. `CH`B. `CH_(2)`C. `CH_(3)`D. The data is not complete |
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Answer» Correct Answer - A `44 g CO_(2)` = 2 mol carbon `3.38 g CO_(2)` = 0.0768 mol carbon `18 g H_(2)O` = 2 mol of hydrogen atoms `0.690 g H_(2)O` = 0.0767 mol hydrogen atom Mole ratio of C:H = 1:1 `:.` Empiriccal formula of hydrocarbon is CH |
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| 336. |
Express of `CO_(2)` is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonated was completely neutralized with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At mass of carbon = 40)A. `200 mL`B. `500 mL`C. `400 mL`D. `300 mL` |
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Answer» Correct Answer - B `Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O` 50 mL of 0.5 mL `Ca(OH_(2)) = (0.5)/(1000)xx50`mol = 0.025 mol 1 mol of `Ca(OH)_(2)` produces `CaCO_(3)=1` mol `:.` 0.025 mol of `Ca(OH)_(2)` will produce `CaCO_(3) = 0.025` mole `CaCO_(3)+2HCl toCaCl_(2)+H_(2)O+CO_(2)` 1 mole of `CaCO_(3)` is neutralized by HCl = 2 mol =0.25 mol `CaCO_(3)` will be neutralized by HCl `=2xx0.025` mol = 0.05 mol 100 mL of 0.1 N HCl contain HCl = 1 eq. =0.1 mol or 0.1 mol of HCl is present in HCl sol = 1000 mL `:.` 0.05 mole of HCl is present in HCl sol `:. (1000)/(0.1)xx0.05 mL = 500 mL` |
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| 337. |
2.5 g of the carbonate of a metal war treated with 100 ml of `1 N H_(2)SO_(4)`. After the completion of the reaction, the solution was boiled off to expel `CO_(2)` and was then titrated against 1 N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20.A. (a)`50`B. (b)`25`C. (c )`75`D. (d)`100` |
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Answer» Correct Answer - d Equivalent weight of metal carbonate `=20+30=50` `2.5 g` of metal carbonate `=2.5/50=0.05 eq.` Number of equivalent of `H_(2)SO_(4)` would have reached `=0.05` Number of equivalent of `H_(2)SO_(4)` taken `=(100xx1)/1000=0.1` `:.` Number of equivalent of `H_(2)SO_(4)` remains unreached `=0.1-0.05=0.05 eq`. `:.` Number of equivalent of alkali consumed `=0.05 eq.` Milli eq. =Normality `xx` Volume in mL. `:. 1.0xxV=0.05xx1000` `V=(0.05xx1000)/1.0=50 mL` |
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| 338. |
An organic compound containing C and H has 92.3 % of carbon, its empirical formula isA. CHB. `CH_(2)`C. `C_(2)H_(2)`D. `CH_(3)` |
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Answer» Correct Answer - A Carbon `= (92.32)/12 = 7.68` Hydrogen `= (7.68)/1 = 7.68` Ratio of `C: H = 1 : 1` `therefore` Empirical formula = CH |
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| 339. |
Compounds having some empirical formula always have sameA. Molecular massB. Molecular formulaC. number of atomsD. percentage composition by mass |
| Answer» d) Compound having same empirical formula always have same percentage composition by mass. | |
| 340. |
What is the equivalent weight of `SnCl_(2)` in the reaction, `SnCl_(2) + Cl_(2) to SnCl_(4)`? (mol.wt. of `SnCl_(2)`=190)A. 95B. 45C. 60D. 30 |
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Answer» a) `SnCl_(2) + Cl_(2) to SnCl_(4)` Eq. wt. of `SnCl_(2)` = Eq. wt of `Cl_(2)` `therefore` (190)/(E_(1)= (71)/(35.5)` or `E_(1) =95` |
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| 341. |
Number of moles of `K_(2)Cr_(2)O_(7)` can be reduced by 1 mole of `Sn^(2+)` ions is:A. `1//3`B. `1//6`C. `2//3`D. 1 |
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Answer» Correct Answer - A `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2 Cr^(3+)+7H_(2)O),((SN^(2+)rarr Sn^(4+)+2e^(-))xx3),(bar(Cr_(2)O_(7)^(2-)+14 H^(+)+3 Sn^(2+) rarr 3Sn^(4+) +2 Cr^(3+)+7 H_(2)O)):}` It is clear from this equation that 3 moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence 1 mol. of `Sn^(2+)` will reduce `1/3` moles of `Cr_(2)O_(7)^(2-)`. |
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| 342. |
Which of the following is not a redox reactionA. `2 Rb+2H_(2)O rarr 2RbOH+H_(2)`B. `2CuI_(2) rarr 2CuI+I_(2)`C. `2H_(2)O_(2) rarr 2H_(2)O+O_(2)`D. `4 KCN+Fe(CN)_(2) rarr K_(4) [Fe(CN)_(6)]` |
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Answer» Correct Answer - D In the reaction `4KCN+Fe(CN)_(2) rarr K_(4)[Fe(CN)_(6)]`, change in oxidation state is not taking place. |
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| 343. |
`2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) rarr 2Z+5O_(2)+8H_(2)O`. In this reaction `Z` isA. `Mn^(+2)`B. `Mn^(+4)`C. `MnO_(2)`D. `Mn` |
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Answer» Correct Answer - A `2 MnO_(4)^(Θ)+5 H_(2)O_(3)+6H^(+) tatt 2Mn^(2+)+5O_(2)+8 H_(2)O`. |
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| 344. |
Assertion :- Equivalent weight of `NH_(3)` in the reaction `N_(2) rarrNH_(3)` is `17//3` while that of `N_(2)` is `28//6`. Reason :- `"Equivalent weight" =("Molecular weight")/("number of "e^(-)" lost or gained/mole")`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 - A Both assertion and reason are true and reason is the correct explanation of assertion. `overset(0)(N_(2))+6e^(-) rarr 2N^(3-)" ":.` Equivalent weight of `NH_(3)=(14+3)/3=17/3 ("M. wt. of "NH_(3))` While for `N_(2)=(14xx2)/6=28/6`. |
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| 345. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Both 12 g of carbon and 27 g of aluminium will have `6.12xx10^(23)` atoms Reason (R) - Gram atomic mass of an element contains Avogadro number of atomsA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 346. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - The atomic mass of an element is not only relative but is average relative mass of an atom Reason (R) - The average word is essential because the element, in general, is a mixture of different isotopes and atomic mass is the average of these relative atomic mass.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 347. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Both 106 g of sodium carbonate and 12 g of graphite have same number of carbon atoms Reason (R) - Both 106 g sodium carbonate and 12 g of graphite contain 1 g-atom of carbonA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 348. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - One mole of water molecules at `4^(@)C` should occupy the volume of 18 mL Reason (R) - Water contains `H_(2)O` molecules bonded by intermolecular H-bonds.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 349. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - The standard unit for expressing the mass of an atom is a.m.u. Reason (R) - a.m.u. is also called avogramA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 350. |
In each of the following two questions two statements are given one labelled as the Assertion(A) or Statement I and the other labelled as the reason (R) or statement II. Examine these statements carefully and mark the correct choice as per following instructions Assertion (A) - Pure water obtained from different sources, such as river, well, spring, sea etc. always contains hydrogen and oxygen in ratio of 1:8 by mass Reason (R) - Mass of reactants and products during chemical or physical changes is always constantA. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |