Question #72f77

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Question #72f77

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Answer 1 · 2015-11-08T20:45:58

$140. g$ $"140. g"$ of potassium chloride and $560. g$ $"560. g"$ of water.

Explanation:

Your starting point here is the percent concentration by mass of the target solution, which you know that must be equal to $20.0 %$ $20.0%$ .

So, what does it mean to have a $20.0 % w/w$ $20.0% "w/w"$ potassium chloride, $KCl$ $"KCl"$ , solution?

As you know, a percent concentration by mass solution is determined by dividing the mass of the solute, which in your case is potassium chloride, by the total mass of the solution, and multiplying the result by $100$ $100$ .

$%w/w = \frac{mass of solute}{mass of solution} \times 100$ $color(blue)("%w/w" = "mass of solute"/"mass of solution" xx 100)$

Now, you know that the total mass of the solution must be $700.0 g$ $"700.0 g"$ . This means that you can rearrange the above equation to solve for what mass of potassium chloride would be needed to make the target solution

$%w/w = \frac{m_{KCl}}{m_{solution}} \times 100 \Rightarrow m_{KCl} = \frac{% w/w \cdot m_{solution}}{100}$ $"%w/w" = m_"KCl"/m_"solution" xx 100 implies m_"KCl" = (%"w/w" * m_"solution")/100$

Plug in your values to get

$m_{KCl} = \frac{20.0 \cdot 700.0 g}{100} = 140. g \to$ $m_"KCl" = (20.0 * "700.0 g")/100 = "140. g" ->$ rounded to three sig figs

So, if you need $140. g$ $"140. g"$ of potassium chloride, and the solution only contains potassium chloride and water, it follows that you must add

$m_{solution} = m_{water} = m_{KCl}$ $m_"solution" = m_"water" = m_"KCl"$

$m_{water} = 700.0 g - 140. g = 560. g$ $m_"water" = "700.0 g" - "140. g" = "560. g"$

of water to make your $20.0 % w/w$ $20.0%"w/w"$ potassium chloride solution.

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Question #72f77

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