Question:medium

A mafic magma with 'Sr' concentration of 400 ppm has undergone 60% fractional crystallization. If 'Sr' is incompatible in the mineral assemblage that crystallizes from the magma with a bulk distribution coefficient of 0.2, the concentration of 'Sr' in the residual liquid is ppm (rounded off to nearest integer).

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Use the Rayleigh fractionation equation C_L = C_0 times F to the power (D-1), where F is the fraction of melt still remaining, not the fraction crystallized.
Updated On: Jul 20, 2026
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Correct Answer: 833

Solution and Explanation

Step 1: Understand why Sr builds up in the melt.
Sr has a bulk distribution coefficient of only 0.2, well below 1. That means the crystallizing minerals take in far less Sr than the melt holds, so every bit of crystal growth leaves the remaining liquid a little richer in Sr. This behavior during ongoing crystal removal is captured by the Rayleigh law
\[ \frac{C_L}{C_0} = F^{D-1} \]

Step 2: Fix the fraction of melt left.
Fractional crystallization of 60% means 60% of the magma mass has turned into crystals and left the system, so the melt still in liquid form is
\[ F = 1 - 0.6 = 0.4 \]

Step 3: Plug the numbers into the ratio.
With $C_0 = 400$ ppm and $D = 0.2$,
\[ \frac{C_L}{400} = (0.4)^{0.2 - 1} = (0.4)^{-0.8} \]

Step 4: Work out $(0.4)^{-0.8}$ carefully.
Write it as $1/(0.4)^{0.8}$. Using $(0.4)^{0.8} = e^{0.8\ln 0.4}$ and $\ln 0.4 \approx -0.9163$, the exponent is $-0.733$, so
\[ (0.4)^{0.8} \approx e^{-0.733} \approx 0.4805 \]
\[ (0.4)^{-0.8} \approx \frac{1}{0.4805} \approx 2.081 \]

Step 5: Multiply back by $C_0$.
\[ C_L \approx 400 \times 2.081 \approx 832.5\ \text{ppm} \]

Step 6: Round and state the answer.
To the nearest whole number, the residual liquid carries about 833 ppm of Sr, an increase from the starting 400 ppm because Sr is left behind by the crystallizing minerals.
\[ \boxed{833} \]
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