What is the molarity of a solution made by dissolving 29.5 g of sodium hydroxide (naoh) in enough water to make 2.45 l of solution? molar mass na: 22.99 g/molmolar mass o: 16.00 g/molmolar mass h: 1.008 g/mol

molarity is defined as the number of moles of solute in 1 L of solution.
the number of moles of NaOH = mass present / molar mass of NaOH
                                                       
molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol
                                                                     
therefore NaOH moles = 29.5 g / 40.00 g/mol = 0.738 mol
 
the number of NaOH moles in 2.45 L solution - 0.738 mol
therefore number of NaOH moles in 1 L = 0.738 mol / 2.45 L = 0.301 mol/L molarity of NaOH = 0.301 M

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