Q.2.25 P - Chemistry: Molecular Nature Of Matter And Change

Question

A compound of iodine and cesium contains 63.94 g of metal and 61.06 g of nonmetal. How many grams of cesium are in38.77 g of compound? How many grams of iodine?

Step-by-Step Solution

Answer

Here the metal is Cesium and Non-metal is Iodine.

Compound= Ce(Cesium) + I(Iodine)

The mass ratio= is 0.511 and the mass I calculated is 18.94g.

Step1. Calculation of the mass ratio

Here the metal is Cesium and Non-metal is Iodine.

Compound= Ce(Cesium) + I(Iodine)

According to the question mass of Ce = 63.94g and Mass of I = 61.06g, Now put these values in the following equation and get the mass ratio as follows:

Mass ratio = $\frac{\mathbf{63}\mathbf{.}\mathbf{94}\mathbf{g}}{\mathbf{125}\mathbf{g}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{511}$

Step2. Calculation of grams of Iodine.

Where m(Ce) is the mass of Cerium, mI is the mass of Iodine, m(Compound) is the mass of the compound, and m(Cesium) is the mass of Cesium.

$$\begin{gathered} \mathit{m}\left(Ce\right)\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{77}\mathit{g}\left(Compound\right)\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{511}\frac{\mathbf{C}\mathbf{e}}{\mathbf{C}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{o}\mathbf{u}\mathbf{n}\mathbf{d}}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{83}\mathit{g}\mathbf{ }\mathit{C}\mathit{e} \\ \mathit{m}\mathit{l}\mathbf{=}\mathit{m}\left(Compound\right)\mathbf{-}\mathit{m}\left(cesium\right)\mathbf{=}\mathbf{38}\mathbf{.}\mathbf{77}\mathit{g}\mathbf{-}\mathbf{19}\mathbf{.}\mathbf{83}\mathit{g}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{94}\mathit{g}\left(l\right) \end{gathered}$$

Hence the mass of Iodine calculated is 18.94g.