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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.
| 1751. |
For an atom 'X', K, L and M shells are completely filled . How many electrons will be present in it ? |
| Answer» SOLUTION :`K - 2 , L - 8 , M - 18 , 28` ELECTRONS - Nickel | |
| 1752. |
For a reaction, the instantaneous rate of reaction with respect to x, y and z is expressed as r=-(1)/(2)(Delta[x])/(Delta t)=-(1)/(3)(Delta[y])/(Delta t)=(1)/(2)(Delta [z])/(Delta t). What would be the stoichiometric equation for the reaction ? |
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Answer» Solution :(i) RELATIONBETWEEN concentration of products, REACTANTS and `K_(c )` (ii) relation between rates of a reaction and stability of the reactants (III) relation between `K_(c )` and stability of reactants (iv) relation between `K_(c )` and RATE of reaction |
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| 1753. |
For ""_(15)^(31)P is one of the isotopes of phosphorus. State the number of protons, neutrons andelectrons in it respectively. |
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Answer» 15, 31, 15 Z = p = `BAR(e)` = 15 A = p + n `THEREFORE` n = A - p = 31 - 15 = 16 |
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| 1754. |
Calculate the number of particles in each of the following: (i) 46 g of Na atoms (number from mass) (ii) 8g O_(2) molecules (number of molecules from mass) (iii) 0.1 mnole of carbon atoms (number from given moles) |
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Answer» Solution :The number of `"ATOMS" // "molecules"= ("GIVEN MASS")/("molar mass") xx "Avogadro number"` i.e., `N= (m)/(M) xx N_(A)` (i) `N= (46)/(23) xx 6.022 xx 10^(23)` [Molar mass of NA = 23 g] `= 2 xx 6.022 xx 10^(23) ~~ 12.044 xx 10^(23)` (ii) `N=(8)/(32) xx 6.022 xx 10^(23)` [Molar mass of `O_(2)` = 32 g] `=(1)/(4) xx 6.022 xx 10^(23) ~~ 1.51 xx 10^(23)` (iii) The number of atoms = `"number of moles" xx "Avogadro number"` `:.` the number of carbon atoms = `0.1 xx 6.022 xx 10^(23) = 6.022 xx 10^(22)` |
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| 1755. |
Calculate the number of particles in each of the following: (i) 46 g of Na atoms (number from mass) (ii) 8g O_(2) molecules (number of molecules from mass) (iii) 0.1 mole of carbon atoms (number from given moles) |
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Answer» Solution :(i) The number of ATOMS = `("given mass")/("MOLAR mass") XX "Avogadro number"` `implies N= (m)/(M) xx N_(0)` `implies N= (46)/(23) xx 6.022 xx 10^(23)` `implies N = 12.044 xx 10^(23)` (II) The number of molecules= `("given mass")/("molar mass") xx "Avogadro number"` `implies N= (m)/(M) xx N_(0)` Atomic mass of oxygen = 16 u `:.` molar mass of `O_(2)` molecules = `16 xx 2` = 32 g `implies N=(8)/(32) xx 6.022 xx 10^(23)` `implies N = 1.5055 xx 10^(23) ~~ 1.51 xx 10^(23)` (iii) The number of particles (atom) = `"number of moles of particles" xx "Avogadro number"` `N = n xx N_(0)` `= 0.1 xx 6.022 xx 10^(23)` `= 6.022 xx 10^(22)` |
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| 1756. |
Calculate the number of particle in 8 g O_(2) molecules (number of molecules from mass) |
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Answer» Solution :The number of molecules = `("given mass")/("molar mass") xx ` Avogadro number `IMPLIES N = (m)/(M) xx N_(0)` Atomic mass of oxygen = 16 u `therefore` Molar mass of `O_(2)` molecules = `16 xx 2 = 32` G `implies N = (8)/(32) xx 6 *022 xx 10^(23)` `= 1*51 xx 10^(23)` |
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| 1757. |
Calculate the number of particle in 46 g of Na atoms (number from mass) |
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Answer» Solution :The number of atoms = `("given mass")/("molar mass") xx` Avogadro number `implies N = (m)/(M) xx N_(0)` `implies N = (46)/(23) xx 6 * 022 xx 10^(23)` `implies N = 12* 044 xx 10^(23)` |
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| 1758. |
Calculate the number of particle in 0.1 mole of carbon atoms (number from given moles) |
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Answer» SOLUTION :The NUMBER of particles (ATOM) = number of moles of particles `xx` Avogadro number `N = n xx N_(0) = 0*1 xx 6*022 xx 10^(23)` `= 6*022 xx 10^(23)` |
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| 1759. |
Calculate the number of neutrons present in the nucleus of an element X which is represented as .^(31)X_(15). |
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Answer» Solution :`.^(31)X_(15)` INDICATE that No FO protons = 15 and mass NUMBER = 31 Mass number = No. of proton + No. of Neutron = 31 No of neutron = 31 - number of protons `= 31-15=16` |
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| 1760. |
Calculate the number of neutrons in the following atoms : (a)""_(27)^(13)AI (b)""_(31)^(15)P ( c) ""_(190)^(76)O (d)""_(54)^(24)Cr |
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Answer» Solution :NUMBER of neutrons (n) = Mass number (A) - Atomic number (z) (a)`""_(27)^(13) `AINumber of neutrons (n) `= 27 (A) - 13 (z) = 14` (b)`""_(31)^(15)P`Number of neuttons (n) `= 31 (A) - 15 (z) = 16` ( c) `""_(190)^(76)O` Number of neutrons (n) `= 190 (A) - 76 (z) = 114` |
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| 1761. |
Calculate the number of moles in the following : (i) 128 g of oxygen (ii) 68 g of ammonia |
| Answer» SOLUTION :(i) 8 , (II) 4 | |
| 1762. |
Calculate the number of moles in 5.75 g of sodium. ("Atomic mass of sodium" = 23) |
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Answer» Solution :1 mol of sodium ATOMS = gram - ATOMIC mass of sodium = 23 g 23 g of sodium = `(5.75)/(23)` mol of sodium = 0.25 mol . |
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| 1763. |
Calculate the number of moles for the following: (i) 52 g of He (finding mole from mass) (ii) 12.044 xx 10^(23) number of He atoms (finding mole from number of particles). |
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Answer» Solution :No. of MOLES = n Given mass = m MOLAR mass =M Given NUMBER of particles = N Avogadro number of particles = `N_(0)` (i) Atomic mass of He = 4 u Molar mass of He = 4g Thus, the number of moles `= ("given mass")/("molar mass") implies n=(m)/(M)=(52)/(4) = 13` mole (II) 1 mole = `6.022 xx 10^(23)` The number of moles = `("given number of particles")/("Avogadro number")` `implies n=(N)/(N_(0))=(12.044 xx 10^(23))/(6.022 xx 10^(23))` = 2 mole |
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| 1764. |
Calculate the number of moles for 12.044 xx 10^(23) number of He atoms (finding mole from number of particles) . |
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Answer» Solution :No. of MOLES = n Given mass = m Molar mass = M Given NUMBER of particles = N Avagadro number of particles = `N_(0)` We know , 1 mole = `6.022 xx 10^(23)` The number of moles = `("given number of particles")/("Avogadro number")` `implies n = (N)/(N_(0)) = (12.044 xx 10^(23))/(6.022 xx 10^(23)) = 2` |
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| 1765. |
Calculate the number of moles for 52 g of He (finding mole from mass) |
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Answer» SOLUTION :No. of MOLES = n Given MASS = m Molar mass = M Given number of PARTICLES = N Avagadro number of particles = `N_(0)` Atomic mass of He = 4 u Molar mass of He = 4 g Thus , the number of moles = `("given mass")/("molar mass")` `implies n = (m)/(M) = (52)/(4) = 13` |
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| 1766. |
Calculate the number of molecules of sulphur (S_(8)) present in 16 g of solid sulphur . |
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Answer» SOLUTION :`(16 XX 6 * 022 xx 10^(23))/(256)` `= 3*76 xx 10^(23) ` molecules |
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| 1767. |
Calculate the number of molecules of sulphur (S_(8)) present in 16 g of solid sulphur. |
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Answer» Solution :Atomic mass of sulphur = 32 G `:.` molar mass of sulphur `(S_(8)) = 32 xx 8` = 256 g Given mass of sulphur = 16 g The number of molecules = `("given mass")/("molar mass") xx "Avogadro number"` `= (16)/(256) xx 6.022 xx 10^(23)` `~~ 0.376 xx 10^(23)` `= 3.76 xx 10^(22)` The number of molecules PRESENT in 16G of sulphur is `3.76 xx 10^(22)` |
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| 1768. |
Calculate the number of molecules of SO_(2) present in 44 g of it. |
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Answer» SOLUTION :The molar MASS of `SO_(2)` = 32 + 32 = 64G The number of molecules = `("GIVEN mass")/("molar mass") xx "Avogadro constant"` `=(44)/(64) xx 6.022 xx 10^(23) = 4.14 xx 10^(23)` |
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| 1769. |
Calculate the number of calcium ions in 5.6 g of calcium oxide . (Ca = 40) |
| Answer» SOLUTION :`6.022 XX 10^(22) G` | |
| 1770. |
Calculate the number of atoms of each element present in 9.8 g of sulphuric acid , H_(2)SO_(4) , (H = 1,S = 32, O = 16) |
| Answer» SOLUTION :`H = 1.2044 XX 10^(22) , S = 6.022 xx 10^(22) , O = 2.4088 xx 10^(22)` | |
| 1771. |
Calculate the number of aluminium ions present in 0.051g of aluminium oxide. |
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Answer» Solution :Molar MASS of aluminium oxide `Al_(2)O_(3) = (2 xx 27) + (3 xx 16)` = 54 + 48 = 102 G The number of molecules `= ("given mass")/("molar mass") xx "Avogadro number"` `= (0.051)/(102) xx 6.022 xx 10^(23)` `= 0.003011 xx 10^(23) ~~ 3.011 10^(20)` 1 molecule of `Al_(2)O_(3)` give `2Al^(3+)` lons `:. 0.051 g Al_(2)O_(3)` gives `2 xx 3.011 xx 10^(20)` `= 6.022 xx 10^(20)` ions |
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| 1772. |
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide . (Hint : The mass of an ion is the same as that of an atom of the same element , Atomic mass of Al = 27 u) |
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Answer» SOLUTION :Mole of aluminium OXIDE `(Al_(2)O_(3))` `=2x27+3x16=102g` i.e., `102g` of `Al_(2)O_(3)=6.022x10^(23)//102x0.051` molecules `=3.011x10^(20)` molecules of `Al_(2)O_(3)` The number of aluminium ion `(Al^(3+))` PRESENT in one molecule of aluminium oxide is 2. Therefore the number of aluminium ions `(Al^(3+))` present in `3.011 x 10^(20)` molecules (0.051g) of aluminium oxide `(Al_(2)O_(3))=2x3.011xx10^(20)` `=6.022x10^(20)` |
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| 1773. |
Calculate the number of aluminium ions in 0.051g of aluminium oxide (Al_(2)O_(3)) |
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Answer» Solution :Step 1: Calcilation of no. of MOLES in 0.0591g of `Al_(2)O_(3)` Gram molecular mass of `Al_(2)O_(3)=2xx"Grams atomic mass of "Al+3xx"Gram atomic mass of O"` `=(2xx27g)+(3xx16g)=102g` Now,a 102g of `Al_(2)O_(3)` =1 MOL `therefore 0.051g of Al_(2)O_(3)=((0.051g))/((102g))xx(1"mol")=0.0005"mol"` Step -II: Calculation of no. of Al ions in 0.001 mole of `Al_(2)O_(3)` 1 mole of `Al_(2)O_(3)` contains Al atoms `=2xxN_(0)` `0.005"mole of"Al_(2)O_(3) "contains Al atoms"=2xx0.005xxN_(0)` `=2xx0.005xx6.022xx10^(23)=6.022xx10^(20)"atoms"` In `Al_(2)O_(3)` valency of Al=3+ No. of aluminium ions (`Al^(3+))` present in the same as the no. of Al atoms. |
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| 1774. |
Calculate the molecule masses of H_(2)O_(2),Cl_(2),CO_(2), CH_(4), C_(2),H_(2),C_(2),H_(4),NH_(3),CH_(3), OH |
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Answer» SOLUTION :Molecule mass of `H_(2)=2` atomic mass of H `=2x1=2u` Molecular mass of `O_(2)=2 x` atomic mass of O `=2xx16=32u` Molecular mass of `Cl_(2)=2x` atomic mass of Cl `=2x35.5=71u` Molecular mass of `Co_(2)=` atomic MASSS of C=2 atomic mass of O `=12+2x 16=44u` Molecules masss of `CH_(4)` : atomic of C=4 atomic mass of H `=12+4x1=16u` Molecules masss of `C_(2)H_(4)=2` x atomic mass of `C+4` x atomic mass of H `=2x 12+4x1=28u` Molecules masss of `NH_(3)`= Atomic mass of `N+3` atomic of mass of H `=14+3x1=17u` Molecules masss of `CH_(3)OH`= Atomic massof C+3 x atomic mass of H+ atomic mass OFO+ Atomic mass of H `=12+3x1+8+1=24u` |
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| 1775. |
Calculate the molecular masses of H_(2), O_(2), Cl_(2), CO_(2), CH_(4), C_(2)H_(6), C_(2)H_(4), NH_(3), CH_(3)OH. |
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Answer» Solution :MOLECULAR mass = `"number of atoms" xx "atomic mass of the element"` The molecular masses of `H_(2) = 2 xx 1 = 2 u""O_(2) = 2 xx 16 = 32 u` `Cl_(2)= 2 xx 35.5 = 71 u""CO_(2)= 1 xx 12 + 2 xx 16 = 12 + 32 = 44 u` `CH_(4)= 1 xx 12+ 4 xx 1 = 12 + 4 = 16 u""C_(2)H_(6) = 2 xx 12 + 6 xx 1 = 24 + 6 = 30 u` `C_(2)H_(4) = 2 xx 12 + 4 xx 1 = 24 + 4 = 28 u""NH_(3) = 1 xx 14 + 3 xx 1 = 14 + 3 = 17 u` `CH_(3)OH= 1 xx 12 +3 xx 1 + 1 xx 16+ 1 xx 1 = 12 + 3 + 16 + 1 = 32 u` |
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| 1776. |
Calculate the molecular mass of the following: NH_(3) |
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Answer» Solution :MOLECULAR mass of `NH_(3) = (1 xx "atomic mass of N") + (3 xx "atomic mass of H")` `= (1 xx 14) + (3 xx 1)` = 14 + 3 = 17 u |
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| 1777. |
Calculate the molecular mass of the following: PCl_(5) |
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Answer» Solution :Molecular mass of `PCI_(5) = (1 xx "ATOMIC mass of P") + (5 xx "atomic mass of Cl")` `= (1 xx 31) + (5 xx 35.5)` = 31 + 177.5 = 208.5 U |
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| 1778. |
Calculate the molecular mass of the following: NaOH |
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Answer» SOLUTION :Molecular mass of NAOH = `(1 xx "atomic mass of NA") + (1 xx "atomic mass of O") + (1 xx "atomic mass of H")` `= (1 xx 23) + (1 xx 16) + (1 xx 1)` = 23 + 16 + 1 = 40 u |
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| 1779. |
Calculate the molecular mass of the following: NaCl |
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Answer» Solution :MOLECULAR mass of NACL = `(1 xx "ATOMIC mass of Na") + (1 xx "atomic mass of Cl")` `= (1 xx 23) + (1 xx 35.5)` = 23 + 35.5 = 58.5 U |
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| 1780. |
Calculate the molecular mass of the following: HCl |
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Answer» Solution :MOLECULAR MASS of HCI = `(1 xx "atomic mass of H") + (1 xx "atomic mass of Cl")` `= (1 xx 1) + (1 xx 35.5)` = 1 + 35.5 = 36.5 U |
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| 1781. |
Calculate the molecular mass of the following: H_(2)O_(2) |
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Answer» Solution :MOLECULAR MASS of `H_(2)O_(2) = (2 xx "atomic mass of H") + (2 xx "atomic mass of O")` `= (2 xx 1) + (2 xx 16)` = 2 + 32 = 34 u |
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| 1782. |
Calculate the molecular mass ofO_(2) |
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Answer» SOLUTION :`O_(2) = 2 xx` ATOMIC mass of oxygen `= 2 xx 16` = `32` |
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| 1783. |
Calculate the molecular mass of NH_(3) |
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Answer» Solution :`NH_(3) = (1 XX ` atomic MASS of Nitrogen) + `( 3 xx ` atomic mass of HYDROGEN) `= (1 xx 14) + (3 xx 1)` `= 14 + 3 = 17` |
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| 1784. |
Calculate the molecular mass ofH_(2) |
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Answer» Solution :`H_(2) = 2 xx` ATOMIC MASS of Hydrogen `=2 xx 1` = 2 |
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| 1785. |
Calculate the molecular mass of CO_(2) |
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Answer» SOLUTION :`CO_(2) = (1 xx` ATOMIC mass of carbon ) + 2 (atomic mass of oxygen) `= (1 xx 12) + (2 xx 16)` = `12 + 32 = 44` |
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| 1786. |
Calculate the molecular mass ofCl_(2) |
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Answer» SOLUTION :`Cl_(2) = 2 XX` ATOMIC mass of CHLORINE `= 2 xx 35.5` `= 71` |
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| 1787. |
Calculate the molecular mass of CH_(4) |
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Answer» SOLUTION :`CH_(4) = (1 xx` atomic mass of CARBON ) + `4 xx` atomic mass of hydrogen `= (1 xx 12) + (4 xx 1) = 12 + 4 = 16` |
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| 1788. |
Calculate the molecular mass ofCH_(3)OH |
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Answer» Solution :`CH_(3)OH = (1 XX 12) + (H xx 4) + (0 xx 10)` `= 12 + 4 + 16 = 32` |
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| 1789. |
Calculate the molecular mass ofC_(2)H_(8) |
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Answer» Solution :`C_(2)H_(8) = ( 2 xx ` atomic MASS of carbon ) + `(8 xx ` atomic mass of HYDROGEN ) `= ( 2 xx 12 ) + ( 8 xx 1)` `= 24 + 8 = 32` |
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| 1790. |
Calculate the molecular mass ofC_(2) H_(4) |
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Answer» SOLUTION :`C_(2)H_(4) = ( 2 xx` ATOMIC mass of CARBON ) + (`4 xx ` atomic mass of hydrogen ) `= ( 2 xx 12) + (4 xx 1)` `= 24 + 4 = 28` |
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| 1791. |
Calculate the Molecular formula unit mass of Na_(2)O, given atomic masses of Na = 234 and O = 164 |
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Answer» SOLUTION :`Na_(2)O = (2 xx NA) + (1 xx 0)` `= (2 xx 23) + (1 xx 16)` `= 46 + 12 = 62` |
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| 1792. |
Calculate the Molecular formula unit mass of K^(2) CO_(3), given atomic masses of K = 394 and O = 164 |
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Answer» SOLUTION :`K_(2) CO_(3) = (2 xx 1C) + (1 xx C) + (3 xx 0) = (2 xx 39) + (1 xx 12) + (3 xx 16)` `= 78 + 12 + 48` =`90 + 48 = 138` |
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| 1793. |
Calculate the molarity of 30 % (w/w) NaOH solution, if the density of the solution is 1.05 g/cc. |
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Answer» SOLUTION :(i) CALCULATION of the volume of 100 g of the solution (II) calculation of number of moles of `NaOH` present in the CALCULATED volume of the solution (iii) 7.88 M |
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| 1794. |
Calculate the molar mass of the following substances : (a) Ethyne, CH_(2) (b) Sulphur molecule, S_(8) (c) Phosphorus molecule, C_(4) (atomic mass of Phosphorus = 31) (d) Nitric acid, HNO_(3) Hydrochloric acid, HCl |
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Answer» SOLUTION :Ethyne, CH Molar MASS of ethyne, `C_(2)H_(2) = 2x12+2x1` `=26g` (b) SULPHUR molecule `S_(8)` Molar mass of Sulphur molecule, `S_(8)=8x32=256g` (c) Phosphorus molecule `P_(4)` `P_(4)=4x31` `=124g` (d) Hydrochloric acid, HCl Molar mass of Nitric acid `HNO_(3)+1+14+3x16=63g` |
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| 1795. |
Calculate the molar mass of the following substances : (a) Ethyne, C_(2)H_(2) (b) Sulphur molecule, S_(8) (c) Phosphorus molecule, P_(4) (Atomic mass of phosphorus = 31) (d) Hydrochloric acid, HCl (e) Nitric acid, HNO_(3). |
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Answer» <P> Solution :The molar mass (Unit, 9) of the following SUBSTANCES are -(a) Ethyne, `C_(2)H_(2) = (2 xx 12) + (2 xx 1)` = 24 +2 = 26 g (b) Sulphur molecule, `S_(8) = 8 xx 32 = 256 g` (c) Phosphorus molecule, `P_(4) = 4 xx 31 = 124 g` (d) HYDROCHLORIC acid, HCl = `(1 xx 1) + (1 xx 35.5)` = 1 + 35.5 = 36.5 g (e) Nitric acid, `HNO_(3) = (1 xx 1) + (1 xx 14) + (3 xx 16)` = 1 + 14 + 48 = 63 g |
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| 1796. |
Calculate the molar mass of Sulphur molecule , S_(i) |
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Answer» SOLUTION :`S_(8) = (SXX 32)` `= 8 xx 32 = 256` |
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| 1797. |
Calculate the molar mass of Nitric acid , HNO_(3) |
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Answer» Solution :`HNO_(3)` `(1 xx 4) + (1 xx N) + (3xx 0)` `(1 xx 1) + (1 xx 14) + (3 xx 16)` `1 + 14 + 48 = 63` |
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| 1798. |
Calculate the molar mass of Hydrochloric acid , HCl |
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Answer» SOLUTION :`HCl = ( 1 XX 4) + (1 xx 4) ` `= ( 1 xx 1) + ( 1 xx 35 - 5)` `= 1 xx 35.5 = 36.5` |
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| 1799. |
Calculate the molar mass of Ethyne , C_(2)H_(2) |
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Answer» Solution :`C_(2)H_(2) = (2 XX C) + (2 xx 4)` `= (2 xx 12) + ( 2 xx 1) ` `= 24 + 2 = 26` |
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| 1800. |
Calculate the molar mass of Phosphorus molecule , P_(4) (Atomic mass of phosphorus = 31) |
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Answer» SOLUTION :`P_(4) = ( 4 XX p)` `= 4 xx 31 = 124` |
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