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Neporo4naja [7]
3 years ago
12

6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm

Hg. Find it's molar mass (show your work)
Chemistry
1 answer:
Stels [109]3 years ago
5 0

<u>Answer:</u> The molar mass of the gas is 35.87 g/mol.

<u>Explanation:</u>

To calculate the mass of gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of gas = 945 mmHg

V = Volume of the gas = 0.35 L

m = Mass of gas = 0.527 g

M = Molar mass of gas = ? g/mo

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of gas =  88^oC=[88+273]=361K

Putting values in above equation, we get:

945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol

Hence, the molar mass of the gas is 35.87 g/mol.

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<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

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

\Delta T_f=iK_fm

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\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

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M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

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<u>Answer:</u> The acceleration of the object is 2m/s^2. If net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

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Mathematically,

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