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2 years ago

How many moles of gas occupy 98L at a pressure of 2.8atm and a temperature of 292k? and what law would you use?

1 answer:

algol132 years ago

8 0

Let's assume that the gas has ideal gas behavior.

Then we can use ideal gas equation,
PV = nRT

Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)

The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?

By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
n = 11.45 mol

Hence,moles of gas is 11.45 mol.

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Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

$N=n\times N_A$

Where : N = Number of atoms

n = moles

$N_A=6.022\times 10^{23}$ = Avogadro number

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Atom percentage of silver = 44.9%

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Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol × 44.9 = 4,843.363 g/mol

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Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

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Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol × 8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol

Mass percentage of gold :

$\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=33.35 \%$

Mass percentage of silver:

$\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=62.80\%$

Mass percentage of copper :

$\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=3.85\%$

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