The Total solution for NCERT class 6-12
Answer - 1 : - (i) Mass of an electron = 9.1 × 10-28 g9.1 × 10-28 g is the mass of = 1 electron(ii) One mole ofelectrons = 6.022 × 1023 electronsMass of 1 electron = 9.1 × 10-31 kgMass of 6.022 × 1023 electrons = (9.1 × 10.31kg) × (6.022 × 1023)= 5.48 × 10-7 kgCharge on one electron = 1.602 × 10-19 coulombCharge on one mole electrons = 1.602 × 10-19 × 6.022 × 1023 =9.65 × 104 coulombs
(ii) One mole ofelectrons = 6.022 × 1023 electronsMass of 1 electron = 9.1 × 10-31 kgMass of 6.022 × 1023 electrons = (9.1 × 10.31kg) × (6.022 × 1023)= 5.48 × 10-7 kgCharge on one electron = 1.602 × 10-19 coulombCharge on one mole electrons = 1.602 × 10-19 × 6.022 × 1023 =9.65 × 104 coulombs
Answer - 2 : -
(i) One mole ofmethane (CH4) has molecules = 6.022 × 1023No. of electrons present in one molecule of CH4 = 6 + 4 = 10No. of electrons present in 6.022 × 1023 molecules of CH4 =6.022 × 1023 × 10= 6.022 × 1024 electrons
Step II. Calculation oftotal number and tatal mass of neutronsNo. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8No. of neutrons present in 3-011 × 1020 atoms (C-14) of carbon= 3.011 × 1020 × 8= 2.408 × 1021 neutronsMass of one neutron = 1.675 × 10-27 kgMass of 2.408 × 1021 neutrons = (1.675X10-27 kg)× 2.408 × 1021= 4.033 × 10-6 kg.
Step II. Calculation oftotal number and mass of protonsNo. of protons present in one molecule of NH3 = 7 + 3 = 10 .No. of protons present in 12.044 × 1020 molecules of NH3 =12.044 × 1020 × 10= 1.2044 × 1022 protonsMass of one proton = 1.67 × 10-27 kgMass of 1.2044 × 1022 protons = (1.67 × 10-27 kg)× 1.2044 × 1022= 2.01 × 10-5 kg.No, the answer will not change upon changing the temperature and pressurebecause only the number of protons and mass of protons are involved.
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Answer - 6 : - (i) Energy of photon (E) = hvh = 6.626 × 10-34 J s ; v = 3 × 1015 Hz = 3 ×1015s-1∴ E = (6.626 × 10-34 Js) × (3 × 1015 s-1) = 1.986 × 1018 JEnergy of photon (E) = hv = hcλh = 6.626 × 10 34 J s; c = 3 × 108 m s-1 ;λ= 0.50 Å = 0.5 × 10-10 m.
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Answer - 8 : - Energy of photon (E) = hcλh = 6.626 × 10-34 Js, c = 3 × 108 m s-1,λ = 4000 pm = 4000 × 10-12 = 4 × 10-9 m
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