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3 years ago

If you have a density of 3g/meander a mass of 15g what is the volume?

Chemistry

1 answer:

erma4kov [3.2K]3 years ago

7 0

M=D*V,
V=m/D
m=15 g
D=3 g/ml
V=15 g/3 g/ml=5 ml

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A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$ where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$

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