The weight of piece of gold, alloyed with silver, is $0.02 k g f$ in air and $0.0187 k g f$ in water. If relative densities of gold and silver are respectively $19.3 a n d 10.5,$ how much gold is there in the piece?

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Solution

Step 1: Given data,
The weight of the alloy in the air = $0.02 k g f$
The weight of the alloy in the water = $0.0187 k g f$
We know that the density of water = $1000 k g / m^{3}$
The relative density of gold = $19.3$
So, the density of gold will be = $19300 k g / m^{3}$
The relative density of silver = $10.5$
So, the density of silver will be = $10500 k g / m^{3}$
Step 2: Finding the volume of the alloy,
$w e i g h t o f w a t e r d i s p l a c e d = w e i g h t o f t h e a l l o y i n a i r - w e i g h t o f a l l o y i n w a t e r = 0.02 k g f - 0.0187 k g f = 0.0013 k g f A l s o, d e n s i t y o f w a t e r = \frac{m a s s o f w a t e r d i s p l a c e d}{v o l u m e o f w a t e r d i s p l a c e d} S o, v o l u m e o f w a t e r d i s p l a c e d = \frac{m a s s o f w a t e r d i s p l a c e d}{d e n s i t y o f w a t e r} = \frac{0.0013 k g}{1000 k g / m^{3}} = 1.3 \times 10^{- 6} m^{3}$
Hence, the volume of the alloy is equal to the volume of water displaced is $1.3 \times 10^{- 6} m^{3}$ .
Step 3: Finding the amount of gold in the alloy,
Let the mass of gold be = $m k g$
So, the mass of silver will be = $(0.02 - m) k g$
$D e n s i t y o f g o l d = \frac{m a s s o f g o l d}{v o l u m e o f g o l d} S o, v o l u m e o f g o l d = \frac{m a s s o f g o l d}{d e n s i t y o f g o l d} = \frac{m k g}{19300 k g / m^{3}} D e n s i t y o f s i l v e r = \frac{m a s s o f s i l v e r}{v o l u m e o f s i l v e r} S o, v o l u m e o f s i l v e r = \frac{m a s s o f s i l v e r}{d e n s i t y o f s i l v e r} = \frac{(0.02 - m) k g}{10500 k g / m^{3}} V o l u m e o f a l l o y = v o l u m e o f g o l d + v o l u m e o f s i l v e r 1.3 \times 10^{- 6} = \frac{m k g}{19300 k g / m^{3}} + \frac{(0.02 - m) k g}{10500 k g / m^{3}} S o, m = (1.3 \times 10^{- 6} - \frac{0.02}{10500}) \times (\frac{19300 \times 10500}{10500 - 19300}) = 0.0139 k g$
Hence, the amount of gold in the alloy is $0.0139 k g$ .

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