(a) What is the mass percentage of iodine in a solution containing $0.035 \mathrm{~mol} \mathrm{I}_{2}\( in \)125 \mathrm{~g}\( of \)\mathrm{CCl}_{4} ?$ (b) Seawater contains \(0.0079 \mathrm{~g}\) of \(\mathrm{Sr}^{2+}\) 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 7.11%, and the concentration of Sr2+ in seawater is 7.9 ppm.

Step by step solution

01

Calculate the molar mass of iodine (#I_2#)

Using the periodic table, we find that the atomic mass of iodine (I) is approximately 127 g/mol. Since there are two iodine atoms in a molecule of iodine, the molar mass of iodine (I2) is 2 * 127 = 254 g/mol.
02

Convert moles of iodine into mass

We know that the solution contains 0.035 mol of I2. To find the mass, we can use the formula: mass of iodine = moles * molar mass. So, mass of iodine = 0.035 mol * 254 g/mol = 8.89 g.
03

Calculate the mass percentage of iodine

The mass of the solution is 125 g, and the mass of iodine in the solution is 8.89 g. To find the mass percentage, we can use the formula: mass percentage = (mass of iodine / mass of solution) * 100 So, mass percentage = (8.89 g / 125 g) * 100 = 7.11% b) Finding the concentration of Sr2+ in seawater in ppm:
04

Convert the mass of Sr2+ per mass of seawater into the mass of Sr2+ per million mass of seawater

We know that seawater contains 0.0079 g of Sr2+ per kilogram of water. To find the concentration of Sr2+ in ppm, we can use the formula: ppm concentration = (mass of solute / mass of solution) * 10^6 So, ppm concentration = (0.0079 g / 1000 g) * 10^6 = 7.9 ppm The mass percentage of iodine in the solution is 7.11%, and the concentration of Sr2+ in seawater is 7.9 ppm.

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