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Anastaziya [24]
3 years ago
14

If 0.089 grams of KI is dissolved in 500g of H2O, what is the concentration of the resulting solution in parts per million?

Chemistry
1 answer:
frez [133]3 years ago
6 0

Answer:

The concentration of the resulting solution in parts per million is 177.97

Explanation:

Parts per million (ppm), is a unit of measure for concentration that refers to the number of units of the substance per million units of the set.

The concentration in parts per million expressed in mass / mass is calculated by dividing the mass of the solute (ms) by the mass of the solution (md, sum of the mass of the solute and the mass of the solvent), both expressed in the same unit and multiplied by 10⁶ (1 million).

ppm=\frac{ms}{md} *10^{6}

So, being:

  • ms: 0.089 grams of KI
  • md: 0.089 grams of KI + 500 grams of H₂O= 500.089 grams

Replacing:

ppm=\frac{0.089 grams}{500.089 grams}*10^{6}

ppm= 177.97

<u><em>The concentration of the resulting solution in parts per million is 177.97</em></u>

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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
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Explanation:

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To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

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Answer:

see explanation

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