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Solved: Calculate the number of cations and anions in each

Chapter 3, Problem 121P

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QUESTION:

Calculate the number of cations and anions in each of the following compounds:

(a) \(0.764 g\) of \(CsI\),

(b) \(72.8\ \text{g}\ K_2Cr_2O_7\),

(c) \(6.54\ \text{g}\ \text{of}\ Hg_2(NO_3)_2\).

ANSWER:

Step 1 of 3

Here, we are asked to calculate the number of cations and anions in the given compounds

(a) Given: 0.764 g of \(CsI\)

First, let’s find the number of moles of \(CsI\):

Molar mass of CsI (Caesium iodide) = 259.81 g/mol

We know 1 mole of any unit contains Avogadro’s number of particles. Here, 1 mole of the \(CsI\) element contains Avogadro’s number of ions, .

Therefore, number of ions will be:

     =

     =  ions

Cs (Caesium) is a metal and hence these ions are the cations. Hence, there are   ions.

The compound, \(CsI\) contains 1 atom of Cs and 1 atom of I (Iodine). A compound always contains an equal number of cations and anions in the neutral state. Hence, \(CsI\) has   anions, i.e   ions.

 

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