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Neporo4naja [7]
4 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]4 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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If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction
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Answer:

<u></u>

  • <u>0.456M</u>

Explanation:

<u>1. Balanced molecular equation</u>

     2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O

<u>2. Mole ratio</u>

     \dfrac{2molHNO_3}{1molBa(OH)_2}

<u>3. Moles of HNO₃</u>

  • Number of moles = Molarity × Volume in liters
  • n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

<u>4. Moles Ba(OH)₂</u>

  • n = 0.700M × 0.0310 liter = 0.0217 mol

<u>5. Limiting reactant</u>

Actual ratio:

   \dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

<u>6. Final molarity of Ba(OH)₂</u>

  • Molarity = number of moles / volume in liters
  • Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
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3 years ago
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