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Chapter 13: Problem 40

(a) What is the mass percentage of iodine in a solution containing 0.035 \(\mathrm{mol}_{2}\) in 125 \(\mathrm{g}\) of \(\mathrm{CCl}_{4} ?\) (b) Seawater contains 0.0079 \(\mathrm{g}\) of Sr \(^{2+}\) per kilogram of water. What is the concentration of \(\mathrm{Sr}^{2+}\) in ppm?

Short Answer

Expert verified
The mass percentage of iodine in the solution is 6.64%, and the concentration of Sr²⁺ ions in seawater is 7.9 ppm.

Step by step solution

01

(Part a: Calculate the mass of iodine)

First, we need to calculate the mass of iodine in grams given the number of moles (0.035 moles). The molecular weight of iodine is approximately 254 g/mol. So, to find the mass of iodine in grams, we use the formula: mass (g) = number of moles x molecular weight mass (g) = 0.035 mol x 254 g/mol mass (g) = 8.89 g So, the mass of iodine in the solution is 8.89 g.
02

(Part a: Calculate the mass percentage of iodine)

Now that we know the mass of iodine, we can find the mass percentage of iodine in the solution using the formula: mass percentage = (mass of solute / mass of solution) x 100 The solution is made up of iodine (8.89 g) and CCl4 (125 g). Therefore, the mass of the solution is: mass of solution = 8.89 g (iodine) + 125 g (CCl4) = 133.89 g Now we can calculate the mass percentage: mass percentage = (8.89 g / 133.89 g) x 100 = 6.64% So, the mass percentage of iodine in the solution is 6.64%.
03

(Part b: Convert grams per kilogram to ppm)

In this part, we have 0.0079 g of Sr²⁺ ions per kilogram of seawater. To convert it to ppm (parts per million), we can use the following conversion: 1 g/kg = 1,000 ppm So, we can multiply the concentration in g/kg by 1,000 to find the concentration in ppm: concentration of Sr²⁺ (ppm) = 0.0079 g/kg x 1,000 = 7.9 ppm Thus, the concentration of Sr²⁺ ions in seawater is 7.9 ppm.

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Most popular questions from this chapter

The Henry's law constant for helium gas in water at \(30^{\circ} \mathrm{C}\) is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and the constant for \(\mathrm{N}_{2}\) at \(30^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) . If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas.

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