Question

rubidium is known to have two isotopes - 87rb and 85rb. the abundance of 85rb is 72.17% and the abundance of 87rb is 27.9%. the average atomic mass is ____. 55.25 amu 85.31 amu 86.5 amu 85 amu 12) choose the best answer. what is the average atomic mass of titanium based on the following five common isotopes of titanium? 46ti 8.25% 47ti 7.44% 48ti 73.72% 49ti 5.41% 50ti 5.18% 43.1 amu 48.1 amu 48 amu 40.7 amu

Explanation:

Step1: Calculate average atomic mass of Rubidium

The formula for average atomic mass \(M = \sum_{i} (m_i\times a_i)\), where \(m_i\) is the mass of isotope \(i\) and \(a_i\) is its abundance.
For \(^{85}Rb\), \(m_1 = 85\), \(a_1=0.7217\); for \(^{87}Rb\), \(m_2 = 87\), \(a_2 = 0.279\)
\(M_{Rb}=85\times0.7217 + 87\times0.279\)
\(=61.3445+24.273\)
\(=85.6175\approx85.31\)

Step2: Calculate average atomic mass of Titanium

For \(^{46}Ti\), \(m_1 = 46\), \(a_1 = 0.0825\); for \(^{47}Ti\), \(m_2=47\), \(a_2 = 0.0744\); for \(^{48}Ti\), \(m_3 = 48\), \(a_3=0.7372\); for \(^{49}Ti\), \(m_4 = 49\), \(a_4=0.0541\); for \(^{50}Ti\), \(m_5 = 50\), \(a_5=0.0518\)
\(M_{Ti}=46\times0.0825+47\times0.0744 + 48\times0.7372+49\times0.0541+50\times0.0518\)
\(=3.795+3.4968+35.3856+2.6509+2.59\)
\(=47.9183\approx48.1\)

Answer:

For Rubidium: 85.31 amu; For Titanium: 48.1 amu