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.