Answer:
A. 7.2%
B. 8.6 ppm
Explanation:
<em>Part A</em>
First let's <em>calculate the mass of I₂</em>, 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%
<em>Part B</em>
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 <u>convert it into mg</u>:
- 8.2x10⁻³g *
= 8.2 mg
Now to convert kg of water into L, we use the density of seawater (1050 kg/m³):
<em>Converting density</em>: 1050 *
= 1.05 kg/L
- Volume of one kilogram of water = 1kgWater ÷ 1.05 kg/L = 0.95 L
Finally we<u> calculate the concentration of Sr⁺²</u>:
- 8.2 mg / 0.95 L = 8.6 ppm