Question

An ore contains 1.34% of the mineral argentile. Ag2s, by weight. How many gram of this ore would have
to be processed in order to obtain 1.00g of pure silver. (Ag)-
(A) 74.6g
(B) 85.7g
(C) 107.9g
(D) 134.Og​

Answer 1

Answer:

(A)74.6g

Explanation:

Given , weight percentage of mineral argentile ,

$\frac{w}{w} \% = 1.34\%$

Therefore ,

100g of ore has 1.34 Ag2S

So , 1g of Ag2S has

$\frac{100}{1.34} \: gram \: \: of \: ore$

$\frac{100}{1.34} \\ \\ = 74.626 \\ \\ = 74.6$

Therefore , 74.6 gram of this ore would have to be processed in order to obtain 1.00 gram of pure silver

Answer 2

Answer:

Option B => 85.7 g

Explanation:

Molar mass of Ag2S :

=>  108 x 2 + 32  = 248 g/mol

Mass of silver in 1 mole of Ag2S :

=> 108 x 2 = 216g

216 g of Ag is contained in 248g of Ag2S

1 g of Ag is consists in = $\dfrac{1 \times 248}{216}$

 => 1.148 g of Ag2S

The ore consists 1.34% Ag2S

∴ 1.34% ≡ 1.148 g

100% :

100% ≡ $\dfrac{100}{1.34 \times 1.148}$

=> 85.7 g.

$\textbf {\underline{\underline{Hence,The option "B" is correct}}}$