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Chemistry 15 Online
OpenStudy (anonymous):

hai guys,can anybody help me? how many grams of HNO3 are required to prepare 500 ml of a 0.601 M HNO3 solution?

OpenStudy (zpupster):

find the moles of HNo3 molarity of HNO3 = Moles of HNo3/ liters=.601M is given moles HNO3 = Molarity* Volume in Liters (.5 L given) lastly grams of HNO3= Moles HNO3* molarmass(g/mol) of HNO3

OpenStudy (anonymous):

mass of HNO3 = ? V = 0.5L M.mass of HNO3 = 63.0067u Molarity = 0.601 molarity = no of moles of solute/volume of solution in L 0.601 = no of moles of solute/0.5 no of moles of solute = 0.3005 mass of HNO3 = no of moles * M.mass = 0.3005*63.0067 = 18.933g

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