1. Calculate the molarity of a 29% (by mass) KOH solution having a density of 1.28 g/mL
2. How many mL of a 69% (by mass) HNO3 solution, having a density of 1.416 g/mL, are required to make 1000 mL of 2.00 M HN03 solution?
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1 What is 29% of 1.28g? Answer 0.3712g So KOH is 0.3712 g/ml
How many moles of KOH is this? 0.3712g KOH/56.1g (molar mass of KOH)
So you have 0.0066 moles KOH per ml times 1000 (go from ml to l)=6.6 moles per litre 6.6 molar
#2 69% of 1.416 is 0.977g HNO3 has a molar mass of 63.0g so it is 0.0155 moles/ml or 15.5 molar
M1V1=M2V2 and you need to solve for V1 (1000ml*2M)/15.5M=129ml of the 15.5 M HNO#