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xxTIMURxx [149]
3 years ago
7

How many grams of chlorine gas are in a 17.8 liter sample at 1.1 atmospheres and 29 degrees Celsius

Chemistry
1 answer:
jasenka [17]3 years ago
5 0
For this problem, we can use the Ideal Gas Law to solve for the number of moles of chlorine gas present in this container, and then we can convert this molar measurement into grams.

The Ideal Gas Law is presented as:

PV=nRT

where P is pressure, V is volume, in liters, n is the number of moles of substance, R is the gas constant associated with the units for pressure, and T is temperature, in Kelvins.

Starting with the IGL, we can substitute the information given in the problem, and then we can simplify to a number of moles of Chlorine gas.

PV=nRT
(1.1 atm)(17.8 L) = n(0.0821 L•atm/mol•K)(302.15 K)
19.58=24.807n
n=0.7893 moles

Chlorine is a gas in this problem, which means it is the di-Chlorine form most commonly found as a gas (Cl2). This molecule weighs twice as much as elemental chlorine (2 mol x 35.453 g/mol). So, we can multiply this new weight by the number of moles of chlorine gas we calculated:

70.9 g/mol • 0.7893 mol = 55.96 g Cl2

55.96 grams of diChlorine gas is present in the sample.
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A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X
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  • <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>

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The relevant fact here is:

  • the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

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  • 84. g solute / 600 ml solution = y / 100 ml solution

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

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<u>The answer is 14. g of solute per 100 ml of solution.</u>

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