A gold crown adulterated with silver was found to weigh 0.540 kg in air and 0.498 kg in water. If the densities of gold and silver are 13900"kgm"^(-3) and 10500"kgm"^(-3), then calculate the mass of silver mixed with gold.

A gold crown adulterated with silver was found to weigh 0.540 kg in ai

1.	A gold crown adulterated with silver was found to weigh 0.540 kg in air and 0.498 kg in water. If the densities of gold and silver are 13900"kgm"^(-3) and 10500"kgm"^(-3), then calculate the mass of silver mixed with gold.
Answer» Solution :Mass of pure gold =(V-X) 13 900 Mass of pure silver =10500 x but `(rho)/(rho_(W))=(W_("air"))/(W_("air)-W_("water"))=(0.540)/(0.540-0.498)=12.857` Density of the IMPURE gold crown `=12.857 xx 1000 =12857 kgm^(-3)` Density of the crown substance `=("mass of gold+mass of silver")/("volume")` 12857V=(V-x)13900+10500x 12857V=13900V-3400x `therefore x/V=(1043)/(Vrho)=(0.3068 xx 10500)/(12857)` `therefore M_(Ag)=0.2505M` i.e, Mass of silver `=0.2505 XX 0.540 =0.1353kg` and the REST pure gold `=0.5400-0.1353=0.4047kg` Percentage of mass of Ag `=(0.1353)/(0.540) xx 100=25.5%` Percentage of mass of Au is 75.5%.