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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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Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

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Temperature of water = 20°C

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The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

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