Question

Carbon tetrachloride at 20 degrees celcius has relative permittivity of 2.24 and density of
1.60 g / cm 3
. Its molecular weight is 156. Calculate the dipole moment of a single molecule of the substance when it is in an electric field of
10 7 V / m

Answer 1

Question #59737, Physics / Electromagnetism

Carbon tetrachloride at 20 degrees Celsius has relative permittivity of 2.24 and density of $1.60\ \mathrm{g/cm^3}$. Its molecular weight is 156. Calculate the dipole moment of a single molecule of the substance when it is in an electric field of $10^7\ \mathrm{V/m}$.

$$\mathrm{CCl_4}$$ $$T = 20{}^\circ\mathrm{C}$$ $$\varepsilon = 2,24$$ $$\rho = 1,6\ \mathrm{g/cm^3} = 1600\ \mathrm{kg/m^3}$$ $$M = 156 \cdot 10^{-3}\ \mathrm{kg/mol}$$ $$E = 10^7\ \mathrm{V/m}$$ $$\mu\text{-?}$$

Solution

The induced electric moments of liquid molecules are the same for all molecules. The induced moment proportional to the field strength acting on the molecule $\mu = \alpha E$, where $\alpha$ - polarizability of the molecule. In accordance with the Clausius - Mossotti for nonpolar gases and liquids, we get the formula: $\frac{(\varepsilon - 1)M}{(\varepsilon + 2)\rho} = \frac{4}{3}\pi N_A \alpha$, consequently the dipole moment of a single molecule of the substance $\mu = \frac{3(\varepsilon - 1)M \cdot E}{4\pi(\varepsilon + 2)\rho N_A}$.

$$\mu = \frac{3 \cdot 1,24 \cdot 156 \cdot 10^{-3}\ \mathrm{kg/mol} \cdot 10^7\ \mathrm{V/m}}{4 \cdot 3,14 \cdot 4,24 \cdot 1600\ \mathrm{kg/m^3} \cdot 6,022 \cdot 10^{23}\ \mathrm{mol^{-1}}} = 11,3 \cdot 10^{-23}\ \mathrm{C/m}.$$

Answer the questions: $\mu = 11,3 \cdot 10^{-23}\ \mathrm{C/m}$.

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