(a) Calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in a solution containing 10.6 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 483 \(\mathrm{g}\) of water. (b) An ore contains 2.86 \(\mathrm{g}\) of silver per ton of ore. What is the concentration of silver in ppm?

Short Answer

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(a) The mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in the solution = 2.15%. (b) The concentration of silver in the ore = 2.86 ppm.

Step by step solution

01

Part (a): Calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

To calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in the solution, we need to divide the mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) by the total mass of the solution and then multiply by 100. The total mass of solution = mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) + mass of water. The given values are: Mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) = 10.6 g Mass of water = 483 g Step 1: Calculate the total mass of the solution Total mass = mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) + mass of water Total mass = 10.6 \(\mathrm{g}\) + 483 \(\mathrm{g}\) = 493.6 \(\mathrm{g}\) Step 2: Calculate the mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) Mass percentage = (mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)/total mass) × 100 Mass percentage = (10.6/493.6) × 100 Mass percentage = 2.15% The mass percentage of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in the solution = 2.15%
02

Part (b): Calculate the concentration of silver in ppm

To determine the concentration of silver in ppm (parts per million), we need to convert the given values into a ratio of mass of silver per million mass units of ore, which is (mass of silver/mass of ore) × 10^6. Given values: Mass of silver = 2.86 g Mass of ore = 1 ton = 1000 kg × 1000 g/kg = 10^6 g Step 1: Calculate the mass ratio of silver to ore Mass ratio = mass of silver/mass of ore Mass ratio = 2.86 / 10^6 Mass ratio = 2.86 × 10^(-6) Step 2: Calculate the concentration of silver in ppm Concentration (ppm) = mass ratio × 10^6 Concentration (ppm) = 2.86 × 10^(-6) × 10^6 Concentration (ppm) = 2.86 The concentration of silver in the ore = 2.86 ppm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Concentration
Understanding solution concentration is essential in chemistry and various practical situations, including cooking, and medicine. Concentration indicates how much of a substance, known as the solute, is contained within a certain volume or mass of another substance, called the solvent. When we refer to the mass percentage of a solute in a solution, we are indicating the fraction of the solution's total mass that is composed of the solute.

For instance, if you dissolve a small amount of salt in water, the salt is the solute, and the water is the solvent. By calculating the mass percentage, you can determine the strength of the solution. A higher mass percentage means a stronger solution, while a lower mass percentage means a more diluted one. This is essential for ensuring correct dosages in pharmaceuticals or the desired taste in food preparations. To calculate the mass percentage, you divide the mass of the solute by the total mass of the solution (which includes both the solute and the solvent) and then multiply by 100.
Parts Per Million (ppm)
The term 'parts per million' (ppm) is another way to express very dilute concentrations of substances. One ppm signifies that one unit of the solute is present in one million units of the total solution. This unit is very convenient for measuring extremely low concentrations, such as impurities in water, air quality, or, as in this exercise's context, the amount of a precious metal within an ore.

Calculating ppm is straightforward when you have the mass of the solute and the mass of the solution. You simply divide the mass of the solute by the total mass of the solution and then multiply the result by one million. This conversion is especially useful when dealing with environmental regulations, where maximum permissible levels of contaminants are often defined in ppm, thus requiring precision and an understanding of how very small amounts relate to larger quantities.
Converting Mass Units
In different regions and for various applications, certain mass units may be preferred. For example, the gram is frequently used in scientific contexts, while the pound may be more common in everyday life in the United States. Converting between these units is essential for accurate communication of measurements. The process of converting units requires an understanding of the equivalent values between units. For instance, understanding that one kilogram is equivalent to 1000 grams is necessary for such conversions.

In chemistry and physics problems, you often need to convert between mass units to carry out calculations correctly. A ton, commonly used in measuring large quantities of raw materials like ore, equals 1000 kilograms or 1 million grams. Therefore, knowing how to convert tons to grams is critical when calculating ppm for substances, such as the concentration of silver in ore, as seen in the provided exercise. This conversion ensures that all measurements are in the same units, which is a fundamental step in performing accurate calculations.

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Most popular questions from this chapter

A dilute aqueous solution of an organic compound soluble in water is formed by dissolving 2.35 g of the compound in water to form 0.250 L of solution. The resulting solution has an osmotic pressure of 0.605 atm at \(25^{\circ} \mathrm{C}\) . Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

Compounds like sodium stearate, called "surfactants" in general, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cme depends on the substance, the solvent, and the temperature. At and above the cmc, the properties of the solution vary drastically. (a) The turbidity (the amount of light scattering) of solutions increases dramatically at the cmc. Suggest an explanation. (b) The ionic conductivity of the solution dramatically changes at the cmc. Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the cmc.

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

List the following aqueous solutions in order of increasing boiling point: 0.120 \(\mathrm{m}\) glucose, 0.050 \(\mathrm{m}\) LiBr, 0.050 \(\mathrm{m}\) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\)

Most fish need at least 4 ppm dissolved \(\mathrm{O}_{2}\) in water for survival. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain 4 \(\mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C} ?\) (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(1.71 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) -atm.)

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