Question

The olivine series of minerals consists of crystals in which Fe and Mg ions may substitute for each other. The density of forsterite $\left( M{{g}_{2}}Si{{O}_{4}} \right)$ is 3.3 g/ml and that of fayalite $\left( F{{e}_{2}}Si{{O}_{4}} \right)$ is 4.2 g/ml. What is the percentage of fayalite in an olivine with a density of 3.9 g/ml?

Answer 1

Hint:The olivine is the series of minerals which consists of $F{{e}^{2+}}$ and $M{{g}^{2+}}$ ions substitute in a crystal lattice. Those ions are present in the form of forsterite and fayalite at different densities.
This problem can be simply solved by the techniques of general percentage solving process.

Complete step-by-step answer:Let us solve the given problem;
The density of forsterite i.e. $M{{g}_{2}}Si{{O}_{4}}$ is 3.3 g/ml and that of fayalite i.e. $F{{e}_{2}}Si{{O}_{4}}$ is 4.2 g/ml.
The density of olivine is 3.9 g/ml.
Now, let us consider;
Forsterite be present in x% and thus, fayalite be present in (100 - x)%; then,
$
 \dfrac{x\times 3.2+\left( 100-x \right)\times 4.2}{100}=3.9 \\
 \Rightarrow x=30% \\
$
Thus, forsterite is 30% in the olivine. Therefore, fayalite will be 70% by the density.

Note:Do note that general percentage calculations can help solve this; do not complicate by the words given. Just concentrate on the numbers described and keep solving.
To solve this problem, even not having any pre knowledge of olivine can help.