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2 years ago

I have 0.725 liters of a 1.500m solution of cacl2. How many grams of cacl2 are in the solution

Chemistry

1 answer:

Stella [2.4K]2 years ago

3 0

Answer:

120.7 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).

∴ mass of CaCl₂ = (M)(V of the solution (L))(molar mass) = (1.5 M)(0.725 L)(110.98 g/mol) = 120.7 g.

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2. 90 mg

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1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

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2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

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Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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