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Paul [167]
3 years ago
11

If 255 g of water has 10.0 g of NaCl dissolved into initially, how much NaCl must be added in order to raise the mass percent of

NaCl by 10%?
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0
M(H₂O) = 255 g.
m₁(NaCl) = 10.0 g.
m₁(solution) = 255 g + 10 g = 265 g.
ω₁ = 10 g / 265 g · 100%.
ω₁ = 3.77% ÷ 100% = 0.0377.
ω₂= 10% ÷ 100% = 0.1.
ω₂= m₁(NaCl) + m₂(NaCl) / m₁(solution) + m₂(NaCl).
0.1 = 10 g + m₂(NaCl) / 265 g + m₂(NaCl).
26.5 g + 0.1·m₂(NaCl) = 10 g + m₂(NaCl).
0.9·m₂(NaCl) = 16.5 g.
m₂(NaCl) = 18.33 g.
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We know that we want a 0.1M solution in 100 mL, so we can plug those into the equation to find the number of moles required:

0.1M= \frac{moles}{0.1L}
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So now we know that the number of moles of the Kool-Aid powder is 0.01. We also know that the chemical formula for this powder is C_{12}  H_{22}  O_{11}.

We can find the molar weight of this powder by taking the atomic weight from the periodic table and multiply it by the number of atoms in the compound. Let's find the molar weight of the powder:

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Then we have 22 H's, each having an atomic weight of 1.008g:

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And finally there are 11 O's, each with an atomic weight of 15.99g:

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Then we add up all of the weights:

144.12g+22.18g+175.89g=342.19g  -> So we know that the molar weight of the Kool-Aid powder is 342.19g.

Now we can find the amount of grams that we need for the desired solution as follows:

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So now we know that we need 0.01 moles of the powder, and we need 3.42g of the powder to make the desired solution.
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