Suppose you need to prepare 1.0 L of Stock solution which must have a concentrat

Question

Suppose you need to prepare 1.0 L of Stock solution which must have a concentration of 150 mg N/L. You start with solid Potassium Nitrate, KNO3 .

A. How many moles of Nitrogen should 1.0 L of Stock solution contain?

B. How many moles of KNO3 should 1.0 L of stock solution contain?

C. How many grams of KNO3 will you have to weigh out to prepare 1.0 L of stock solution? (Show ALL work!) (Reminder: the concentration of stock must be known to 3 significant figures!)

2. Another Stock solution you may have to prepare must contain 80.0 mg Phosphorus /L. Using steps similar to those used for Nitrogen Stock solution in Question 1, find how many grams of Potassuim Dihydrogen Phosphate (KH2PO4 ) you will need to weigh out to prepare 1.0 L of such Stock solution

Explanation / Answer

1.(A): Given the mass of nitrogen in 1L of the stock solution = 150 mg = 150 mg x (1 g / 1000 g) = 0.150 g

Atomic mass of nitrogen = 14.0 g/mol

Hence moles of nitrogen = mass / atomic mass = 0.150 g / 14.0 g/mol = 0.0107 mol (answer)

(B): 1 mole of KNO3 contains 1 mole of nitrogen atom.

Hence 1 L of the stock solution would contain 0.0107 mol of KNO3 (answer)

(C): Moles of KNO3 need to put to prepare 1L stock solution = 0.0107 mol

Molecular mass of KNO3 = 101.1 g/mol

Hence mass of KNO3 need to weigh out =  0.0107 mol x 101.1 g/mol = 1.082 g (answer)

Q.2:

Given the mass of phosphorus in 1L of the stock solution = 80.0 mg = 80.0 mg x (1 g / 1000 g) = 0.080 g

Atomic mass of phosphorus = 31.0 g/mol

Hence moles of phosphorus = mass / atomic mass = 0.080 g / 31.0 g/mol = 0.00258 mol

1 mole of KH2PO4 contains 1 mole of phosphorus atom.

Hence 1 L of the stock solution would contain 0.00258 mol of KH2PO4

Hence moles of KH2PO4 needed to prepare 1L stock solution = 0.00258 mol

Molecular mass of KH2PO4 = 136.08 g/mol

Hence mass of KH2PO4 need to weigh out =  0.00258 mol x 136.08 g/mol = 0.351 g (answer)

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