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hoa [83]
1 year ago
5

which intermolecular forces are responsible for the dissolution of ethylene glycol? select all that apply.

Chemistry
1 answer:
aksik [14]1 year ago
8 0

The intermolecular forces that are responsible for the dissolution of Ethylene glycol in water is hydrogen bonding dipole-dipole forces and dispersion forces.

Both ethylene glycol and water contains the pair of hydrogen and oxygen.

The hydrogen of one atom create a bond with the oxygen of other atom this results in the formation of intra molecular hydrogen bonding.

The electron are non uniformly distributed over the molecule or the atom which results in the fluctuation of the electron density in the atom.

So it creates are dispersion forces which is present all over the molecule this forces helps to increase the strength of the bond formed between the ethylene glycol and water because they have large masses.

Both ethylene glycol and water are polar molecules because of being polar they form dipole and the dipole of both the molecules interact with each other in order to form bond between the atoms which eventually results in the formation dissolution of ethylene glycol in water.

To know more about intermolecular forces, visit,

brainly.com/question/2193457

#SPJ4

Compete Question - which intermolecular forces are responsible for the dissolution of ethylene glycol? select all that apply hydrogen bonding, dipole-dipole, dispersion and Ion dipole interaction.

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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
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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.

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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.

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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}

7 0
3 years ago
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