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Chapter 25: Problem 55

A small quantity of \(\mathrm{NaCl}\) containing radioactive \(_{11}^{24} \mathrm{Na}\) is added to an aqueous solution of \(\mathrm{NaNO}_{3}\). The solution is cooled, and \(\mathrm{NaNO}_{3}\) is crystallized from the solution. Would you expect the \(\mathrm{NaNO}_{3}(\mathrm{s})\) to be radioactive? Explain.

Short Answer

Expert verified
Yes, the resulting \(\mathrm{NaNO}_{3}(\mathrm{s})\) would be radioactive because it includes the radioactive $_{11}^{24} \mathrm{Na}$ isotope from the solution.

Step by step solution

01

Identify the radioactive isotope

In this problem, the radioactive isotope is $_{11}^{24} \mathrm{Na}$. It's important to note that this isotope of sodium behaves chemically identical to other sodium isotopes.
02

Analyze the crystallization process

When the sodium nitrate, \(\mathrm{NaNO}_{3}\), is cooled and crystallized from the solution, all kinds of sodium ions present in the solution (including the radioactive isotope $_{11}^{24} \mathrm{Na}$ and the common sodium isotope $_{11}^{23} \mathrm{Na}$) are included into the crystal structure of the resulting sodium nitrate.
03

Conclude if the resulting solid is radioactive

Since the radioactive isotope of sodium is included into the crystallized sodium nitrate along with non-radioactive isotopes, the resulting sodium nitrate solid, \(\mathrm{NaNO}_{3}(\mathrm{s})\), would be radioactive.

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Most popular questions from this chapter

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