### Solution Found! # 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) $$6.54\ \text{g}\ \text{of}\ Hg_2(NO_3)_2$$.

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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