I have worked this problem out a number of ways, but every answer I come up with is incorrect. I have posted my last answers underneath the question. Please advise.
(a) What is the mass percentage of iodine (I2) in a solution containing 0.060 mol I2 in 105 g of CCl4?
My answer: 0.05711%
(b) Seawater contains 0.0079 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm?
My answer: 0.007900 ppm
I know how you obtained the wrong answer. You divided 0.06/105 and multiplied by 100 to get 0.0561 BUT you don't have the units right.
Atomic mass I = 126.9
molar mass I2 = 126.9 x 2 = 253.8
0.06 mol I2 = (253.8g/mol) x 0.06 mol = 15.22 g I2.
Total mass is 15.22 + 105 = ??
mass percent = (15.22/total mass)*100 = ?? about 13% or so. Check my thinking. Check my work.
I'll post the b part separately.
Do they, (they meaning the question makers) expect you to take into account the density of sea water since it isn't 1.00 g/mL. I will assume you are to consider density of sea water to be the same as fresh water. It won't make much difference in the answer.
Your answer you obtained by 0.0079 g Sr/1,000 g water and that is 0.0079 parts per thousand. Just set up a ratio of (0.0079 g Sr/1000 g H2O) = X g (Sr/1,000,000 g H2O) and solve for X. Check my thinking. Check my work.
(a) To find the mass percentage of iodine (I2) in the solution, we need to calculate the mass of iodine (I2) and divide it by the total mass of the solution, then multiply by 100 to get the percentage.
Given:
- Moles of I2 = 0.060 mol
- Mass of CCl4 = 105 g
Step 1: Calculate the molar mass of I2.
The molar mass of iodine (I) is 126.9 g/mol.
The molar mass of I2 is (2 * 126.9) g/mol = 253.8 g/mol.
Step 2: Calculate the mass of I2 in the solution.
To find the mass of I2, we can use the equation:
mass = moles * molar mass
Mass of I2 = 0.060 mol * 253.8 g/mol = 15.228 g
Step 3: Calculate the mass percentage.
Mass percentage = (mass of I2 / total mass of solution) * 100
Mass percentage = (15.228 g / 105 g) * 100 = 14.50%
Therefore, the correct mass percentage of iodine (I2) in the solution is 14.50%, not 0.05711%.
(b) To find the concentration of Sr2+ measured in ppm (parts per million), we need to calculate the amount of Sr2+ in grams and divide it by the total mass of the solution, then multiply by 1,000,000.
Given:
- Mass of Sr2+ = 0.0079 g
- Mass of water = 1 kg = 1000 g
Step 1: Calculate the amount of Sr2+ in the solution.
Amount of Sr2+ = 0.0079 g
Step 2: Calculate the concentration in ppm.
Concentration (ppm) = (amount of Sr2+ / total mass of solution) * 1,000,000
Concentration (ppm) = (0.0079 g / 1000 g) * 1,000,000 = 7.9 ppm
Therefore, the correct concentration of Sr2+ measured in ppm is 7.9 ppm, not 0.007900 ppm.
To solve these problems, you need to use the appropriate formulas and unit conversions. Let's go through each question step by step:
(a) To find the mass percentage of iodine in the solution, you need to calculate the mass of iodine (I2) and divide it by the total mass of the solution. Here's how to do it:
1. Calculate the molar mass of I2: I2 has an atomic mass of 126.9 g/mol, so the molar mass of I2 is 2 * 126.9 g/mol = 253.8 g/mol.
2. Convert the 0.060 moles of I2 to grams: Multiply the number of moles by the molar mass of I2.
0.060 mol I2 * 253.8 g/mol = 15.228 grams of I2.
3. Calculate the total mass of the solution: Add the mass of iodine (15.228 g) to the mass of the solvent (105 g CCl4).
Total mass = 15.228 g I2 + 105 g CCl4 = 120.228 g.
4. Calculate the mass percentage: Divide the mass of iodine by the total mass and multiply by 100 to get the percentage.
(15.228 g I2 / 120.228 g) * 100 = 12.66%.
Therefore, the correct mass percentage of iodine in the solution is 12.66%, which differs from your answer of 0.05711%.
(b) To calculate the concentration of Sr2+ in seawater measured in ppm (parts per million), you need to convert the given concentration in grams per kilogram to ppm. Here's how to do it:
1. Convert the given concentration of Sr2+ from grams to milligrams: Multiply by 1000.
0.0079 g Sr2+ * 1000 = 7.9 mg Sr2+.
2. Convert the mass of Sr2+ to ppm: Divide the mass of Sr2+ by the mass of the solution and multiply by 10^6.
(7.9 mg Sr2+ / 1 kg of water) * (10^6 mg / 1 kg) = 7.9 ppm.
Hence, the correct concentration of Sr2+ in seawater is 7.9 ppm, which differs from your answer of 0.007900 ppm.
I hope this explanation helps you understand how to solve these problems correctly. Make sure to check your calculations and units throughout the process to avoid errors.