Some Basic Concepts Of Chemistry Question 76
Question: A 25.0 mm $ \times $ 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is $ 19.32,g\text{/}cm^{3} $ . How many gold atoms are in the sheet? (Atomic weight: Au = 197.0)
Options:
A) $ 7.7\times 10^{23} $
B) $ ~1.5\times 10^{23} $
C) $ 4.3\times 10^{21} $
D) $ 1.47\times 10^{22} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Volume of gold foil $ =25\times 40\times 0.25mm^{3} $ $ =250\times {10^{-3}}cm^{3} $ Mass of gold foil $ =19.32\times 250\times {10^{-3}}g $ $ =4.83g $ No. of gold atoms $ =\frac{4.83}{197}\times N _{A} $ $ =1.47\times 10^{22} $
