Question

i) Calculate the number of vibrational degrees of freedom for hydrogen
fluoride and chloroethene.
ii) For a compound, molar extinction coefficient is 215 m^2 mol-1 at 255 nm.
What concentration of the compound in a solution will cause a 30%
decrease in the intensity of 255 nm radiation? The cell thickness is 0.01 m.

Answer 1

Answer on Question #44198, Physics, Other

Question:

i) Calculate the number of vibrational degrees of freedom for hydrogen fluoride and chloroethene.

ii) For a compound, molar extinction coefficient is $215\mathrm{m}^{\wedge}2$ mol-1 at $255\mathrm{nm}$ . What concentration of the compound in a solution will cause a $30\%$ decrease in the intensity of $255\mathrm{nm}$ radiation? The cell thickness is $0.01\mathrm{m}$ .

Answer:

i) Hydrogen fluoride is a chemical compound with the formula HF, so it is linear molecule:

H—F

The degrees of vibrational modes for linear molecules can be calculated using the formula:

$$3 n - 5$$

n is equal to the number of atoms within the molecule of interest.

For HF $n = 2$ :

$$3 \cdot 2 - 5 = 1$$

The molecular formula for chloroethene is $C_2H_3Cl$ :

So, it is nonlinear molecule.

The degrees of freedom for nonlinear molecules can be calculated using the formula:

$$3 n - 6$$

For $C_2H_3Cl$ n=6:

$$3 \cdot 6 - 6 = 12$$

Answer: hydrogen fluoride 1, for chloroethene 12.

ii) The molar absorption coefficient $(\varepsilon)$ is a measurement of how strongly a chemical species absorbs light at a given wavelength. It is an intrinsic property of the species; the actual absorbance, A, of a sample is dependent on the pathlength, $\ell$, and the concentration, c, of the species via the Beer-Lambert law,

$$A = \varepsilon c l$$

From Beer-Lambert law:

$$A = - \ln \left(\frac {I}{I _ {0}}\right)$$

If we have $30\%$ decrease in the intensity:

$$\frac {I}{I _ {0}} = \frac {100 - 30}{100} = 0.7$$ $$A = - \ln 0.7$$

And concentration equals:

$$c = \frac {A}{\varepsilon l} = = - \frac {\ln 0.7}{215 \frac {m ^ {2}}{m o l} 0.01 m} = 0.166 \frac {m o l}{m ^ {3}}$$

Answer: $0.166\frac{mol}{m^3}$

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