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Chapter 8: Problem 3

Metals which have low melting point are refined by liquation.

Short Answer

Expert verified
Question: Explain the process of liquation and its application in refining metals with low melting points. Answer: Liquation is a metallurgical process used to refine and purify metals with low melting points. The process involves heating a mixture of the desired metal and impurities on a sloping hearth or furnace to a temperature slightly above the metal's melting point but below the impurities' melting points. The low melting point metal melts, flows down the slope, and is collected separately from the solid impurities. The purified metal is then cooled and solidified. Further refinement, such as electro-refining, can be applied if needed.

Step by step solution

01

Understanding Liquation

Liquation is a metallurgical process used to separate and purify metals based on their melting points. It is particularly used to refine metals with low melting points. The principle behind liquation is that a metal with a lower melting point will melt when heated, while the impurities or other metals with higher melting points remain solid. This allows the liquid metal to be separated from the solid impurities.
02

Setting up the Liquation Process

To carry out the liquation process, a sloping hearth or a furnace specifically designed for liquation is used. The metal mixture (low melting point metal mixed with impurities) is placed on the sloping hearth, which is heated from below to a temperature slightly above the melting point of the desired metal but below the melting points of the impurities.
03

Melting and Separation of Metals

As the temperature of the furnace or hearth increases, the low-melting-point metal in the mixture starts to melt, while the impurities remain solid. Due to the sloping surface, the molten metal flows down the slope and is collected in a container at the bottom. The solid impurities on the hearth or furnace can be later removed and discarded.
04

Cooling and Solidification

Once the molten metal is separated from the impurities, it needs to be cooled and solidified. This can be done by removing the heat source or transferring the molten metal to a cooler container. As the temperature decreases, the purified metal solidifies into a solid mass.
05

Final Refinement (Optional)

In some cases, a single pass through the liquation process may not be enough to achieve the desired level of purity. If that is the case, the purified metal can be subjected to further liquation or any other purification technique, like the electro-refining process, to achieve a higher degree of purity. Liquation is a simple and effective method for refining and purifying metals with low melting points, allowing the desired metal to be easily separated from impurities based on their respective melting points.

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Most popular questions from this chapter

When a certain amount of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), is burnt completely, \(7.04 \mathrm{~g} \mathrm{CO}_{2}\) is formed. What is the mass of \(\mathrm{H}_{2} \mathrm{O}\) formed, simultaneously? (a) \(1.62 \mathrm{~g}\) (c) \(6.48 \mathrm{~g}\) (c) \(3.24 \mathrm{~g}\) (d) \(2.28 \mathrm{~g}\)

How many grams of \(90 \%\) pure \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) can be produced from \(250 \mathrm{~g}\) of \(95 \%\) pure \(\mathrm{NaCl} ?\) (a) \(640.6 \mathrm{~g}\) (b) \(288.2 \mathrm{~g}\) (c) \(259.4 \mathrm{~g}\) (d) \(320.3 \mathrm{~g}\)

A compound has carbon, hydrogen, and oxygen in \(3: 3: 1\) atomic ratio. If the number of moles in \(1 \mathrm{~g}\) of the compound is \(6.06 \times 10^{-3}\), the molecular formula of the compound will be (a) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{O}_{4}\)

The vapour density of a sample of \(\mathrm{SO}_{3}\) gas is 28 . Its degree of dissociation in to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is (a) \(1 / 7\) (b) \(1 / 6\) (c) \(6 / 7\) (d) \(2 / 5\)

A compound contains elements \(X\) and \(\mathrm{Y}\) in \(1: 4\) mass ratio. If the atomic masses of \(X\) and \(Y\) are in \(1: 2\) ratio, the empirical formula of compound should be (a) \(\mathrm{XY}_{2}\) (b) \(\mathrm{X}_{2} \mathrm{Y}\) (c) \(\mathrm{XY}_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)
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