Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate the atomic mass (average) of chlorine using the following data :
\(\begin{array}{lcl} & \text { % Natural Abundance } & \text { Molar Mass } \\ { }^{35} \mathrm{Cl} & 75.77 & 34.9689 \\ { }^{37} \mathrm{Cl} & 24.23 & 36.9659\end{array}\)
\(\begin{array}{lcl} & \text { % Natural Abundance } & \text { Molar Mass } \\ { }^{35} \mathrm{Cl} & 75.77 & 34.9689 \\ { }^{37} \mathrm{Cl} & 24.23 & 36.9659\end{array}\)
Solution:
2979 Upvotes
Verified Answer
Fractional abundance of \({ }^{35} \mathrm{Cl}=0.7577\)
Molar mass \(=34.9689\)
Fractional abundance of \({ }^{37} \mathrm{Cl}=0.2423\)
Molar mass \(=36.9659\)
Average atomic mass
\(=(0.7577)(34.9689 \mathrm{amu})+(0.2423)(36.9659 \mathrm{amu})\) \(=26.4959+8.9568=35.4527\)
Molar mass \(=34.9689\)
Fractional abundance of \({ }^{37} \mathrm{Cl}=0.2423\)
Molar mass \(=36.9659\)
Average atomic mass
\(=(0.7577)(34.9689 \mathrm{amu})+(0.2423)(36.9659 \mathrm{amu})\) \(=26.4959+8.9568=35.4527\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.