Respuesta :
rate of effusion of phosphorus / rate of effusion of neon = 0,404.
rate of effusion of neon = 1/√M(Ne).
rate of effusion of neon = 1/√20,18 = 0,222.
rate of effusion of phosphorus = rate of effusion of neon · 0,404.
rate of effusion of phosphorus = 0,222 · 0,404 = 0,0897.
rate of effusion of phosphorus = 1/√M(Pₓ).
√M(Pₓ) = 1 ÷ 0,0897 = 11,14.
M(Pₓ) = (11,15)² = 124.
N(P) = 124 g/mol ÷ 31 g/mol = 4.
There are 4 atoms in a molecule of gaseous white phosphorus.
rate of effusion of neon = 1/√M(Ne).
rate of effusion of neon = 1/√20,18 = 0,222.
rate of effusion of phosphorus = rate of effusion of neon · 0,404.
rate of effusion of phosphorus = 0,222 · 0,404 = 0,0897.
rate of effusion of phosphorus = 1/√M(Pₓ).
√M(Pₓ) = 1 ÷ 0,0897 = 11,14.
M(Pₓ) = (11,15)² = 124.
N(P) = 124 g/mol ÷ 31 g/mol = 4.
There are 4 atoms in a molecule of gaseous white phosphorus.
The number of phosphorus atom in white phosphorus is [tex]\boxed4[/tex].
Further Explanation:
Effusion is the process by which molecules of gas travel through a small hole from high pressure to the low pressure. According to the graham’s law, effusion rate of a gas is inversely proportional to the square of the molar mass of gas.
[tex]{\text{Rate of effusion}}\left({\text{R}}\right)\propto\frac{1}{{\sqrt {{\mu}}}}[/tex] .....(1)
Where [tex]\mu[/tex]is the molar mass of a gas.
[tex]{{\text{R}}_{{\text{white phosphorus}}}}\propto\frac{1}{{\sqrt{{{\text{\mu}}_{{\text{white phosphorus}}}}} }}[/tex] ...... (2)
And, the rate of effusion of neon gas is expressed as
[tex]{{\text{R}}_{{\text{Neon}}}} \propto \frac{1}{{\sqrt {{{\tex{\mu }}_{{\text{Neon}}}}} }}[/tex] ......(3)
Since rate of effusion of white phosphorus is 0.404 times of rate of effusion of neon. Therefore,
[tex]{{\text{R}}_{{\text{white phosphorus}}}}=0.404{{\text{R}}_{{\text{Neon}}}}[/tex] ......(4)
Substitute the value of [tex]{{\text{R}}_{{\text{white phosphorus}}}}[/tex] and [tex]{{\text{R}}_{{\text{Neon}}}}[/tex] from equation (2) and equation (3) respectively in equation (4).
[tex]\frac{1}{{\sqrt {{\mu _{{\text{white phosphorus}}}}}}}=0.4041\sqrt{{\mu _{{\text{Neon}}}}}[/tex] ......(5)
Simplify equation (5) as follows:
[tex]\begin{aligned}\dfrac{1}{\sqrt{\mu_{\text{White Phosphorous}}}}&=0.404\left(\dfrac{1}{\sqrt{\mu_{\text{Neon}}}}\right)\\\sqrt{\mu_{\text{White Phosphorous}}}&=\dfrac{\sqrt{\mu_{\text{Neon}}}}{0.404}\\\mu_{\text{White Phosphorous}}&=\left(\dfrac{\sqrt{\mu_{\text{Neon}}}}{0.404}\right)^{2}\\&=\dfrac{\mu_{\text{Neon}}}{(0.404)^{2}}\end{aligned}[/tex]
Therefore the molar mass of white phosphorus is,
[tex]{{{\mu}}_{{{white phosphorus}}}} = \frac{{{{{\mu}}_{{{Neon}}}}}}{{{{\left({0.404} \right)}^2}}}[/tex] ......(6)
The molar mass of neon is 20.18 g/mol.
Substitute the value of molar mass of neon in the equation number (6).
[tex]\begin{aligned}{{t{\mu }}_{{\text{white phosphorus}}}}=\frac{{20.18{\text{g/mol}}}}{{{{\left( {0.404} \right)}^2}}}\\=123.6{\text{ g/mol}}\\\end{gathered}[/tex]
Therefore the molar mass of white phosphorus is 123.6 g/mol.
Formula to calculate the number of phosphorus atoms present in white phosphorus is,
[tex]{\text{Number of P atom}} = \frac{{{\text{molar mass of white phosphorus}}}}{{{\text{molar mass of P}}}}[/tex] ......(7)
The atomic mass of phosphorus atom is 30.97 g/mol.
Substitute molar mass of white phosphorus and atomic mass of phosphorus in equation (7).
[tex]\begin{aligned}{\text{Number of P atom}}&= \frac{{{\text{123}}{\text{.6 g/mol}}}}{{{\text{30}}{\text{.97 g/mol}}}}\\&= 3.99\\&\approx4{\text{ atoms}}\\\end{aligned}[/tex]
Therefore the number of phosphorus atoms in white phosphorus is 4.
Learn more:
1. Balanced chemical equation https://brainly.com/question/1405182
2. Calculation of volume of gas: https://brainly.com/question/3636135
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ideal gas equation
Keywords: Effusion, rate of effusion, white phosphorus, neon, molar mass, atomic mass, number of phosphorus atoms, 4 atoms, rate of effusion of white phosphorus, rate of effusion of neon, molar mass of neon.
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