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Answer 1 · 2017-10-16T18:42:31

$25 g$ $25 g$ of $N a_{2} S O_{4}$ $Na_2SO_4$ is required.

$C % = \frac{mass of solute}{mass of solute + mass of solvent}$ $C % = ("mass of solute") /("mass of solute"+"mass of solvent")$

Let, $\to$ $rarr$ mass of solute $(N a_{2} S O_{4}) = x$ $(Na_2SO_4) = x$
$\to$ $rarr$ mass of solvent (water) $= 100 g$ $= 100g$

$C % = \frac{x}{x + 100} \cdot 100 % = 20 %$ $C %= x/(x+100)*100%=20%$

Hence the equation becomes:

$x = (x + 100) \cdot 0.2$ $x=(x+100)*0.2$

$x = 0.2 x + 20$ $x=0.2x+20$

$0.8 x = 20$ $0.8x=20$

$x = \frac{20}{0.8}$ $x=20/0.8$

$x = 25 g$ $x=25g$ . (Answer).

Hope it Helps:)