Question

How many grams of CaCl2 are present in 20.0 mL of a 0.490 M solution

Answer 1

Answer on Question#59127 – Chemistry – Organic chemistry

Question: How many grams of CaCl₂ are present in 20.0 mL of a 0.490 M solution.

Solution:

$$n(CaCl_2) = V(CaCl_2) \cdot C_M(CaCl_2) = \frac{20.0 \, L \cdot 0.490 \, \text{mol}}{1000} = 0.0098 \, \text{mol}$$ $$m(CaCl_2) = n(CaCl_2) \cdot M(CaCl_2) = 0.0098 \, \text{mol} \cdot 111 \, \frac{\text{g}}{\text{mol}} = 1.09 \, \text{g}$$

Answer: 1.09 g CaCl₂

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