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Chapter 10: Problem 105

The melting point of a fictional substance \(X\) is \(225^{\circ} \mathrm{C}\) at \(10.0\) atm. If the density of the solid phase of \(\mathrm{X}\) is \(2.67 \mathrm{~g} / \mathrm{cm}^{3}\) and the density of the liquid phase is \(2.78 \mathrm{~g} / \mathrm{cm}^{3}\) at \(10.0 \mathrm{~atm}\), predict whether the normal melting point of \(X\) will be less than, equal to, or greater than \(225^{\circ} \mathrm{C}\). Explain.

Short Answer

Expert verified
The normal melting point of substance X at 1 atm will be greater than 225℃. This is because substance X contracts upon melting, as the density of the liquid phase is higher than the density of the solid phase. According to the Clausius-Clapeyron equation, a decrease in pressure from 10 atm to 1 atm should lead to an increase in the melting point.

Step by step solution

01

Understand the relationship between melting point, pressure, and density

According to the Clausius-Clapeyron equation, an increase in pressure will lead to an increase in the melting point if the substance expands upon melting (liquid phase density lower than solid phase). In contrast, if the substance contracts upon melting (liquid phase density higher than solid phase), an increase in pressure will lead to a decrease in the melting point.
02

Compare the given densities of substance X

We are given the density of the solid phase of substance X as \(2.67 \frac{g}{cm^3}\), and the density of the liquid phase as \(2.78 \frac{g}{cm^3}\). Comparing these values, we can see that the density of the liquid phase is higher than the density of the solid phase.
03

Predict the effect of changing pressure on the melting point of substance X

Since the density of the liquid phase is higher than the density of the solid phase for substance X, it implies that substance X contracts upon melting. As a result, a decrease in pressure from 10 atm to 1 atm (normal pressure) should lead to an increase in the melting point of substance X.
04

Conclude the prediction

Based on our analysis, we can predict that the normal melting point of substance X at 1 atm will be greater than 225℃. The reason for this is that substance X contracts upon melting, and a decrease in pressure will lead to an increase in the melting point.

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Most popular questions from this chapter

The molar enthalpy of vaporization of water at \(373 \mathrm{~K}\) and \(1.00\) atm is \(40.7 \mathrm{~kJ} / \mathrm{mol}\). What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

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