1.

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 =1.6726xx10^(-27)kg)

Answer»

Solution :a) `" 1 mol of NH"_(3) =" 17 G NH"_(3)= 6.022xx 1023` molecules of`NH_(3) = (6.022xx 10^(23))XX (7 + 3)` protons `= 6.022 xx 10^(24)` protons
In 34 mg, i.e., 0.034 g NH3, protons `=(6.022xx10^(24))/(17)xx0.034=1.2044xx10^(22)`
(b)Mass of one proton `= 1.6726 xx 10^(-27) kg`
Mass of `1.2044xx 10^(22)` protons `= (1.6726 xx 10^(-27)) xx (1.2044 xx10^(22)) kg = 2.0145 xx 10^(-5) kg`


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