Asked Aug 27, 2019

How many grams of ammonium nitrateNH4NO3, must be dissolved to prepare 300. mL of a 0.166 M aqueous solution of the salt? 


Expert Answer

Step 1

The molarity of a solution is given by the equation (1) in which M is the molarity of solution, n is the number of moles of solute and V is the volume of solution in litres.


The volume of solution is 300 mL which is equal to 0.300 L.


The number of moles of ammonium nitrate in the given solution is calculated in the equation (2) by substituting the values of molarity and volume in litres in equation (1).


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(1) M = v(in L) 0.166 mol/L . 0.300 L (2) n 0.0498 mol

Step 2

The number of moles of a substance (n) is given by the equation (3) in which m is the given mass of substance and M is the molar mass.


The mass of ammonium ni...


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(3) n M m 0.0498 molS0.04 g/mol (4) m 3.98 g


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