How many hydrogen atoms are present in \(51.0 g\) of hydrogen sulphide?
\(\left[ H =1, S =32, N _{ A }=6.02 \times 10^{23}\right]\)
A. \(6.02 \times 10^{23}\)
B. \(9.03 \times 10^{23}\)
C. \(9.03 \times 10^{24}\)
D. \(1.81 \times 10^{24}\)
Correct Answer: D
Explanation
Mole of \(H _{2} S =\frac{51.0 g }{34 gmol ^{-1}}=1.5 mol\)
Since 1 mole of \(H _{2} S\) contains 2 moles of hydrogen atoms, mole of hydrogen atoms present in \(1.5\) mole of \(H _{2} S\) equals \(1.5 mol \times 2=3 mol\)
Number of \(H\) atoms \(=\) Mole of \(H\) atom \(\times 6.02 \times 10^{23}\)
\(=3 \times 6.02 \times 10^{23}\)
\(=1.806 \times 10^{24}\)
\(=1.81 \times 10^{24}\)
