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What is the mass percentage of iodine (I2) in a solution containing 4.0×10−2 mol I2 in 130 g of CCl4? Express your answer using two significant figures. % Request Answer Part B Seawater contains 8.2×10−3 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm? Express your answer using two significant figures. ppm p p m Request Answer

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Answer:

A. 7.2%

B. 8.6 ppm

Explanation:

Part A

First let's calculate the mass of I₂, using the known value of moles and the molecular weight (253.8 g/mol)

  • 4.0x10⁻² mol I₂ * 253.8 g/mol = 10.152 g I₂

Mass percentage is calculated using the mass of I₂ and the total mass (mass of I₂ + mass of CCl₄)

  • Total mass = 130 + 10.152 = 140.152 g
  • Mass percentage I₂ = 10.152 / 140.152 * 100 = 7.2%

Part B

The concentration of Sr⁺² in ppm is calculated using the formula

  • mg Sr⁺² / L water

We're given the mass of Sr⁺² in grams, so now we convert it into mg:

  • 8.2x10⁻³g * [tex]\frac{1000mg}{1g}[/tex] = 8.2 mg

Now to convert kg of water into L, we use the density of seawater (1050 kg/m³):

Converting density: 1050 [tex]\frac{kg}{m^{3}}[/tex] *  [tex]\frac{1m^{3}}{1000L}[/tex] = 1.05 kg/L

  • Volume of one kilogram of water = 1kgWater ÷ 1.05 kg/L = 0.95 L

Finally we calculate the concentration of Sr⁺²:

  • 8.2 mg / 0.95 L = 8.6 ppm

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