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mote1985 [20]
3 years ago
15

What is the percentage strength (w/w) of a solution made by dissolving 62.5 g of potassium chloride in 187.5 ml of water?

Chemistry
1 answer:
kiruha [24]3 years ago
8 0

Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.

Given data:

Mass of the solute, potassium chloride = 62.5 g

Volume of water (solution) = 187.5 ml

We know that the density of water = 1 g/ml

Therefore, the mass corresponding to the given volume of water

= 187.5 ml * 1 g/1 ml = 187.5 g

We have a solution of 62.5 g of potassium chloride in 187.5 g water

Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33

The percentage strength = 33.33 %

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A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

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Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

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  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
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Answer:

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2 years ago
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Answer:

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the volume will be calculated using ideal gas equation

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Where

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V = ?

n = moles = 2.31 mol

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T = 25 °C = 298.15 K

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Answer:

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