A student dissolved 3.00 g of Ca(NO₃)₂ in enough water to make 100.mL of stock solution.He took 4.00 mL of the solution then diluted it with water to give 275 mL of a final solution.How many grams of NO₃^- ion are there in the final solution? A)0.329 g B)0.00073 g C)0.0453 g D)0.0906 g

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Quizplus · Accepted Answer

The molar mass of Ca(NO3)2 is 164.1 g/mol. The initial concentration of Ca(NO3)2 is (3.00 g / 164.1 g/mol) / 0.1 L = 0.183 M. Since each formula unit of Ca(NO3)2 provides 2 NO3- ions, the concentration of NO3- ions is 0.366 M. The moles of NO3- in 4.00 mL of the stock solution is 0.366 M * 0.004 L = 0.001464 mol. In the final 275 mL solution, the concentration remains the same, so the mass of NO3- is 0.001464 mol * 62.01 g/mol = 0.0906 g.

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