Question

question 10 of 39
on another planet, the isotopes of titanium have the given natural abundances.

isotope	abundance	mass (u)
$^{48}$ti	12.500%	47.94795
$^{50}$ti	16.100%	49.94479

what is the average atomic mass of titanium on that planet?

average atomic mass: \boxed{} u

Explanation:

Step1: Convert percentages to decimals

For \(^{46}\text{Ti}\): \(71.400\% = 0.714\)
For \(^{48}\text{Ti}\): \(12.500\% = 0.125\)
For \(^{50}\text{Ti}\): \(16.100\% = 0.161\)

Step2: Calculate the contribution of each isotope

Contribution of \(^{46}\text{Ti}\): \(0.714\times45.95263\)
\(0.714\times45.95263 = 32.810\) (approx)

Contribution of \(^{48}\text{Ti}\): \(0.125\times47.94795\)
\(0.125\times47.94795 = 5.993\) (approx)

Contribution of \(^{50}\text{Ti}\): \(0.161\times49.94479\)
\(0.161\times49.94479 = 8.041\) (approx)

Step3: Sum the contributions

Total average atomic mass = \(32.810 + 5.993 + 8.041\)
\(32.810+5.993 = 38.803\)
\(38.803 + 8.041 = 46.844\) (more precise calculation: )

Let's do precise calculation:

Contribution of \(^{46}\text{Ti}\): \(0.714\times45.95263 = 0.714\times45.95263 = 32.81017782\)

Contribution of \(^{48}\text{Ti}\): \(0.125\times47.94795 = 5.99349375\)

Contribution of \(^{50}\text{Ti}\): \(0.161\times49.94479 = 8.04111119\)

Sum: \(32.81017782+5.99349375 + 8.04111119=32.81017782 + 14.03460494=46.84478276\)

Answer:

\(46.84\) (or more precisely \(46.8448\)) u