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Question

Physics Question on Electric Charge

A cup contains 250g250\,g of water. Find the total positive charge present in the cup of water

A

1.34×1019C1.34 \times 10^{19}\, C

B

1.34×107C1.34 \times 10^{7}\, C

C

2.43×1019C2.43 \times 10^{19}\, C

D

2.43×107C2.43 \times 10^{7}\, C

Answer

1.34×107C1.34 \times 10^{7}\, C

Explanation

Solution

Mass of water =250g= 250\, g, Molecular mass of water =18g= 18\,g Number of molecules in 18 g of water =6.02×1023= 6.02 \times 10^{23} Number of molecules in one cup of water =25018×6.02×1023=\frac{250}{18}\times6.02\times10^{23} Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. \therefore Total positive charge present in one cup of water =25018×6.02×1023×10×1.6×1019C=1.34×107C=\frac{250}{18}\times6.02\times10^{23}\times10\times1.6\times10^{-19}\,C=1.34\times10^{7}\,C