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How many L of a 1.1M solution contain 0.303 mol of CaCl_{2}? How many g of NaCl are contained in 0.703 L of a 1.2M solution? How many g of NaCl are contained in 913 mL of a 0.45M solution?

Answer :

1. The solution of 0.303 mol of CaCl₂ requires approx 0.276 L of 1.1M solution.

2. 0.703 L of a 1.2M NaCl solution contains around 100.93 g of NaCl.

3. 913 mL of a 0.45M NaCl solution holds about 36.585 g of NaCl.

In the first reaction, to calculate the volume of a 1.1M CaCl₂ solution containing 0.303 mol of the compound, you can use the formula:

Volume (L) = Number of moles / Concentration (Molarity)

Volume = 0.303 mol / 1.1 M ≈ 0.276 L

For the second reaction, the amount of NaCl in grams in a given volume of solution can be determined using the formula:

Mass (g) = Volume (L) × Concentration (Molarity) × Molar Mass (g/mol)

Molar mass of NaCl = 58.44 g/mol

Mass = 0.703 L × 1.2 M × 58.44 g/mol ≈ 100.93 g

For the third reaction, the calculation is similar:

Mass = 0.913 L × 0.45 M × 58.44 g/mol ≈ 36.585 g

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