The relative atomic mass of boron (which exists in two isotopic forms 10B and 11B) is 10·81. What will be the abundance of 10B and 11B, respectively (consider a sample of 100 atoms) ?
- (a)19% and 81%
- (b)81% and 19%
- (c)38% and 62%
- (d)62% and 38%
Correct — A, 19% and 81%. The relative atomic mass is the abundance-weighted average of the isotope masses. If a fraction x of the atoms are 10B and (1 - x) are 11B, then 10x + 11(1 - x) = 10.81, which simplifies to 11 - x = 10.81, so x = 0.19. That makes 10B about 19% and 11B about 81% of the sample, matching option (a). It also fits that the average (10.81) lies much closer to 11 than to 10, because the heavier isotope dominates.
- (b)81% and 19% — The two abundances are swapped — if 10B were 81%, the average would work out near 10.2, not 10.81.
- (c)38% and 62% — These fractions do not satisfy 10x + 11(1 - x) = 10.81 — they would give an average of about 10.62.
- (d)62% and 38% — These give an average near 10.38, far from the stated relative atomic mass of 10.81.
The relative (average) atomic mass of an element is the mean of its isotope masses weighted by their fractional abundances. Working backwards from the known average lets you solve for the abundances of the isotopes.
The quickest check is that the weighted average sits closer to whichever isotope is more abundant. Since 10.81 is near 11, the 11B isotope must dominate — so the split has to be a small 10B share and a large 11B share.
- Relative atomic mass equals the sum of (isotope mass x fractional abundance) over all isotopes.
- Boron has two stable isotopes, 10B and 11B, and a standard relative atomic mass of about 10.81.
- Solving 10x + 11(1 - x) = 10.81 gives x = 0.19, so 10B is about 19% and 11B about 81%.
- The average always lies nearer the more abundant isotope, which is a useful sanity check.
The average atomic mass (10.81) is close to 11, so the heavier 11B isotope dominates at 81% versus 19%.
- Swapping the two abundances — check which isotope the average lies closer to before choosing.
- Reading the relative atomic mass as if it were the mass number of a single isotope.
A back-calculation item — set up (mass x abundance) = average and solve; the average's proximity to 11 tells you 11B is the major isotope.
No directly related past PYQ was found.
- practice — not a real PYQ
Chlorine (relative atomic mass about 35.5) has the isotopes 35Cl and 37Cl. Their approximate abundances are:
- (a)25% and 75%
- (b)75% and 25%
- (c)50% and 50%
- (d)90% and 10%
Answer(b) 75% and 25% — 35 x 0.75 + 37 x 0.25 = 35.5, the relative atomic mass of chlorine.
- practice — not a real PYQ
The relative atomic mass of an element is best described as:
- (a)the mass of its most abundant isotope
- (b)the sum of protons and neutrons in its nucleus
- (c)the abundance-weighted average of its isotope masses
- (d)the mass of one mole of its atoms
Answer(c) the abundance-weighted average of its isotope masses.