Question

Chapter 21, Problem 033

Calculate the number of coulombs of positive charge in 275 cm³ of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

Answer 1

Water is H2O

So total number of proton in 1 water molecule will be = 2 protons in H + 8 protons in O = 10 protons

Amount of charge on single water molecule = 10*e = 10*1.6*10^-19 = 1.6*10^-18 C

Given that

Amount of water = 275 cm^3

Density of water = 1 g/cm^3

Mass of water = Volume*density = 275*1 = 275 gm

Number of moles of water = Mass/Molecular weight = 275/18 = 15.28 moles

Number of molecules of water = n*Na

Na = Avogadro's number = 6.023*10^23

Number of molecules of water = 15.28*6.023*10^23 = 9.20*10^24

So,

Total amount of positive charge = Number of molecules*Charge on one molecule

Total amount of positive charge = 9.20*10^24*1.6*10^-18 = 14.72*10^6 C

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