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Rashid [163]
3 years ago
12

Determine the molecular mass of a gas where 3.87 g occupies 0.896 l at standard conditions.

Chemistry
1 answer:
Arturiano [62]3 years ago
5 0
Let's assume that the gas has ideal gas behavior.

Ideal gas law,
<span>
PV = nRT               (1)

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.

</span>n = m/M          (2)

Where, n is number of moles, m is mass and M is molar mass.

From (1) and (2),
PV = (m/M) RT

By rearranging,
M = (mRT)/PV              (3)

P =  standard pressure = 1 atm = 101325 pa
V = 0.896 L = 0.896 x 10⁻³ m³
R = 8.314 J mol⁻¹ K⁻¹<span>
T = Standard temperature = 273 K
m = </span>3.87 g = 3.87 x 10⁻³ kg<span> 
M = ?
</span><span>
By appying the formula,
M =(</span>3.87 x 10⁻³ kg x 8.314 J mol⁻¹ K⁻¹ x 273 K) /101325 pa x  0.896 x 10⁻³m³
<span>M = 0.0967 kg
M = 96.7 g.

Hence, the molar mass of the gas is 96.7 g.

</span>
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3 years ago
A sample of Manganese (II) chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is
IRINA_888 [86]

Answer:

no. of water molecules associated to each molecule of MnCl_2 = 4

Explanation:

Mass of MnCl_2 before heating = 19.8 g

Mass of MnCl_2 after heating = 12.6 g

Difference in mass of MnCl_2 before and after heating

                                 = 19.8 - 12.6 = 7.2 g

Difference in mass corresponds to mass of water driven out.

Molar mass of water = 18 g/mol

No. of moles of water = \frac{7.2}{18} = 0.4\ mol

Mass of MnCl_2 obtained after heating is mass of anhydrous MnCl_2.

Mass of anhydrous MnCl_2 = 12.6 g

Molar mass of MnCl_2 = 125.9 g/mol

No. of mol of anhydrous MnCl_2 = \frac{125.9}{125.9} = 0.1\ mol

so,

0.1 mol of MnCl_2 have 0.4 mol of water

1 mol of MnCl_2 will have = \frac{0.4}{0.1} =4\ mol

Hence, no. of water molecules associated to each molecule of MnCl_2 = 4

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3 years ago
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