
Assuming the olivine series of minerals consists of crystal in which \[F{e^{2 + }}\] and \[M{g^{2 + }}\] ions may substitute for each other causing substitutional impurity defect without changing the volume of the unit cell. In the olivine series of minerals, oxide ion exits as fcc with \[S{i^{4 + }}\]occupying \[\dfrac{1}{4}th\] of octahedral void and divalent ion occupying \[\dfrac{1}{4}th\] of the tetrahedral void. The density of Forsterite (Magnesium silicate) is \[3.21\dfrac{g}{{cc}}\] and that of Fayalite (Ferrous Silicate) is\[4.34\dfrac{g}{{cc}}\]. If the density of olivine is\[3.88\dfrac{g}{{cc}}\], then which of the following statements is /are correct?
A. Olivine contains \[40.71\% \]Forsterite and \[59.29\% \]Fayalite.
B. Fayalite is \[F{e_2}Si{O_4}\]and Forsterite is\[MgSi{O_4}\].
C. Percentage of Forsterite\[MgSi{O_4}\]is\[59.29\].
D. Both options A and B.
Answer
513.6k+ views
Hint : Chemical composition of olivine falls somewhere close to pure Forsterite (\[MgSi{O_4}\]) and pure Fayalite(\[F{e_2}Si{O_4}\]). Around there \[Mg\]and can elective uninhibitedly for one another in the mineral’s nuclear design-in any proportion, known as “solid arrangement”.
Complete step-by-step solution:
According to the question we get-
Number of \[{O^{ - 2}}\]atoms
\[
= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} \\
= 1 + 3 \\
= 4 \\
\]
Number of \[S{i^{4 + }}\]atoms \[ = \dfrac{1}{4} \times \]octahedral void
\[
= \dfrac{1}{4} \times 4 \\
= 1 \\
\]
Number of \[M{g^{2 + }}\] atoms \[\dfrac{1}{4} \times \]tetrahedral void
\[
= \dfrac{1}{4} \times 8 \\
= 2 \\
\]
We know the formula of Forsterite \[ = MgSi{O_4}\] and Fayallite\[ = F{e_2}Si{O_4}\]
Let us assume the Forterite is \[x\% \] and Fayalite is \[(100 - x)\% \]
Therefore,
\[
\dfrac{{x \times 3.21 + (100 - x) \times 4.34}}{{100}} = 3.88 \\
\therefore x = 40.71\% - Forsterite \\
\\
\]
And the percentage of Fayalite \[\left( {100 - 40.71} \right) = 59.29\].
Therefore, the right option is D.
Note: Olivine has additionally been utilized as a hard–headed material. It is utilized to make a headstrong block and utilized as projecting sand. It is the primary mineral found on the earth's surface.
Complete step-by-step solution:
According to the question we get-
Number of \[{O^{ - 2}}\]atoms
\[
= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} \\
= 1 + 3 \\
= 4 \\
\]
Number of \[S{i^{4 + }}\]atoms \[ = \dfrac{1}{4} \times \]octahedral void
\[
= \dfrac{1}{4} \times 4 \\
= 1 \\
\]
Number of \[M{g^{2 + }}\] atoms \[\dfrac{1}{4} \times \]tetrahedral void
\[
= \dfrac{1}{4} \times 8 \\
= 2 \\
\]
We know the formula of Forsterite \[ = MgSi{O_4}\] and Fayallite\[ = F{e_2}Si{O_4}\]
Let us assume the Forterite is \[x\% \] and Fayalite is \[(100 - x)\% \]
Therefore,
\[
\dfrac{{x \times 3.21 + (100 - x) \times 4.34}}{{100}} = 3.88 \\
\therefore x = 40.71\% - Forsterite \\
\\
\]
And the percentage of Fayalite \[\left( {100 - 40.71} \right) = 59.29\].
Therefore, the right option is D.
Note: Olivine has additionally been utilized as a hard–headed material. It is utilized to make a headstrong block and utilized as projecting sand. It is the primary mineral found on the earth's surface.
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