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How many grams of KNO₃ are needed to make 250 ml of a solution that is 0.135 M?

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Final answer:

To prepare 250 ml of a 0.135 M solution of KNO₃, 3.41 grams of KNO₃ are needed after calculating the number of moles required and using the molar mass of KNO₃.

Step-by-step explanation:

To calculate the mass of potassium nitrate (KNO₃) required to make 250 ml of a 0.135 M solution, first we need to find the number of moles of KNO₃ needed. This is done by multiplying the molarity (M) of the solution by the volume (L) of the solution. The volume should be in liters, so we convert 250 ml to liters, which is 0.250 L.

Number of moles of KNO₃ = 0.135 M × 0.250 L = 0.03375 mol

To find out the mass of these moles, we multiply the number of moles by the molar mass of KNO₃. The molar mass of KNO₃ is K (39.10 g/mol) + N (14.01 g/mol) + 3×O (3 × 16.00 g/mol) which equals 101.10 g/mol. Thus:

Mass of KNO₃ = 0.03375 mol × 101.10 g/mol = 3.41 g

You will need to weigh out 3.41 grams of KNO₃ to make 250 ml of a 0.135 M solution.

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