1.

(i) Calculate the total number of electrons present in one mole of methane (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of `.^(14)C` (Assume that the mass of neutron `= 1.675 xx 10^(-27) kg`) (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of `NH_(3)` at S.T.P. (Assume the mass of proton `= 1.6726 xx 10^(-27) kg`) Will the answer change if temperature and pressure are changed ?

Answer» (i) 1 molecule of `CH_(4)` contains electrons `= 6 + 4 = 10`
`:.` 1 mole, i.e., `6.022 xx 10^(23)` molecules will contain electrons `= 6.022 xx 10^(24)`
(ii) (a) 1 g atom of `.^(14)C = 14g = 6.022 xx 10^(23)` atom `= (6.022 xx 10^(23)) xx 8` neutrons (as each `.^(14)C` atom has `14 - 6 = 8` neutrons)
Thus, 14g or 14000 mg have `8 xx 6.022 xx 10^(23)` neutrons
`:.` 7 mg will have neutrons `= (8 xx 6.022 xx 10^(23))/(14000) xx 7 = 2.4088 xx 10^(21)`
(b) Mass of 1 neutrons `= 1.67 xx 10^(-27) kg`
`:.` Mass of `2.4088 xx 10^(-21)` neutrons `= (2.4088 xx 10^(21)) (1.67 xx 10^(-27) kg) = 4.0347 xx 10^(-6) kg`
(iii) (a) 1 mol of `NH_(3)` = 17 g `NH_(3) = 6.022 xx 10^(23)` molecules of `NH_(3)`
`= (6.022 xx 10^(23)) xx (7 +3)` protons `= 6.022 xx 10^(24)` protons
`:.` 34 mg i.e., `0.034 g NH_(3) = (6.022 xx 10^(24))/(17) xx 0.034 = 1.2044 xx 10^(22)` protons
(b) Mass of one proton `= 1.6726 xx 10^(-27) kg`
`:.` Mass of `1.2044 xx 10^(22)` protons `= (1.6726 xx 10^(-27)) xx (1.2044 xx 10^(22)) kg = 2.0145 xx 10^(-5) kg`
There is no effect of temperture and pressure.


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