Question

A silver ornament of mass 'm' gram is polished with gold equivalent to1%of the mass of silver. compute the ratio of no. f atoms of gold and silver ornament

Answer 1

Mass of silver = 'm' g   ----------(1)
Mass of gold = m/100g  --------(2)

Number of atoms of silver(Ag) = Mass/Atomic mass * Avagadro's number(NA)
= m/108*NA      ---------------(from(1))
Atomic weight of silver = 108g/mole

No: of atoms of gold(Au) = Mass / Atomic mass * Avagadro's no:(NA)
= m/100 * 197 * NA  ------------from (2)
Ratio of number of atoms of gold to silver = Au:Ag
= m/100 * 197 * NA : m/108 * NA
=108:100 * 197
=108:19700
=1:182

[Thank You]