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Mama L [17]
3 years ago
11

A sample of carbon dioxide gas occuples a volume of 2.5L at standard temperature and pressure (STP). What will be the volume of

a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature?
Chemistry
1 answer:
MArishka [77]3 years ago
5 0

Answer:

The correct answer is 5.0 L

Explanation:

STP are defined as T=273 K and P= 1 atm

By using the ideal gas equation, we can calculate the number of moles (n) of the gas at a volume V=2.5 L:

PV= nRT

⇒n= (PV)/(RT) =(1 atm x 2,5 L)/(0.082 L.atm/K.mol x 273 K)= 0.112 mol

For a sample of argon gas, with the same number of moles (0.112 mol) but twice the temperature (T = 273 K x 2= 546 K):

V= (nRT)/P = (0.112 mol x 0.082 L.atm/K.mol x 546 K)/1 atm = 5.0 L

That is consistent with the fact that when a gas is heated, it expanses. So, if the temperature increases twice, the volume also increases twice.

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
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<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

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Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

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Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

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Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

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