Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A silver ornament of mass’m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

A silver ornament of mass’m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

Answers (1)

Answer:

Mass of silver =mg
Mass of gold deposited =\frac{m\times1}{100}=\frac{m}{100}g
No. of atoms of silver =\frac{Mass}{Atomic mass}\times N_{A}=\frac{m}{108}\times N_{A}
No. of atoms of gold =\frac{Mass}{Atomic mass}\times N_{A}=\frac{m}{100\times 197}\times N_{A}
Ratio of the number of atoms of gold to silver =\frac{m}{100\times 197}\times N_{A}:\frac{m}{108}\times N_{A}
108:100\times 197=108:19700=1.182.41

Posted by

infoexpert26

View full answer

Similar Questions

Latest Question