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Chapter 37: Problem 23

Express the 7.84 -eV ionic cohesive energy of \(\mathrm{NaCl}\) in kilocalories per mole of ions.

Short Answer

Expert verified
The ionic cohesive energy of \(\mathrm{NaCl}\) is approximately 180.47 kcal/mol of ions.

Step by step solution

01

Convert from eV to J

The energy of 7.84 eV is given in electron volts, not joules. To convert to joules, the fact that 1 eV equals \(1.60218 x 10^{-19}\) J is used, thus\[7.84\, eV = 7.84\, (\,eV) * 1.60218 * 10^{-19}\, (\,J / eV) = 1.25611 * 10^{-18}\, J \]
02

Convert from J to calories

Given that 1 J equals 0.239006 kcal, the conversion is follows:\[(1.25611 * 10^{-18} J) * (0.239006 cal / 1 J) = 2.99969906 * 10^{-19}\, cal \]
03

Convert from calories to kcal

Since 1 kcal equals 1000 calories, the conversion to kcal become\[2.99969906 * 10^{-19} cal * (1 kcal / 1000 cal) = 2.99969906 * 10^{-22}\, kcal \]
04

Convert from kcal per ion to kcal per mole

We know that 1 mole contains Avogadro's number of ions which is \(6.022x10^{23}\). So, we multiply \(2.99969906 * 10^{-22}\, kcal\) with Avogadro's number to get kcal per mole\[2.99969906 * 10^{-22}\, kcal * 6.022 * 10^{23}\, mol^-1 = 180.471725 kcal/mol \]

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Most popular questions from this chapter

The Fermi energy in metals is much higher than the thermal energy at typical temperatures. Why does this make the mean speed of conduction electrons nearly independent of temperature?

What fraction of conduction electrons in a metal at absolute zero have energies less than half the Fermi energy?

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