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

Question

espectively 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, KNO 1. A. How many moles of Nitrogen should 1.0 L of Stock solution contain? (Show ALL work!) B. How many moles of KNO, should 1.0 L of stock solution contain? (Show ALL work!) C. How many grams of KNO, 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!)

Explanation / Answer

The required concentration of stock solution = 150 mg N/L

Where N = nitrogen

A. The mass of N present in 1 L of stock solution = 150 mg

= 150/1000 g (Since 1 g = 1000 mg)

= 0.15 g

The molar mass of N = 14 g/mol

Therefore, the no. of moles of N present in 1 L of stock solution = 0.15 g/14 g mol-1

= 0.0107 mol

B. One mole of KNO3 contains 1 mole of N.

i.e. The no. of moles of KNO3 = no. of moles of N in 1 L of stock solution = 0.0107 mol

C. The molar mass of KNO3 = 101.1 g/mol

Therefore, the mass of KNO3 needed to weigh for preparing 1 L of stock solution = 0.0107 mol * 101.1 g/mol

= 1.082 g

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