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0.5mol/lit HNO3Moderator: BioTeam
7 posts • Page 1 of 1
Re: 0.5mol/lit HNO3You need to know what molarity you are starting with. You can then use the formula
M1V1=M2V2. You need to know what starting M1V1 is, and then solve for V2. For example if you started with a 100mL of 1 Molar HNO3, and desired M2 is .5 your formula would be (1M(.1L))/.5 = 200 mL. which means you would add 100 mL H2O to you 1 Molar solution.
Re: 0.5mol/lit HNO3The bottle should also tell you the % composition which you can use, along with the specifc gravity, to calculate the molarity of the solution. The % composition tells you how many grams of substance there is in 100 grams of solution. The specific gravity is in essence the density of the solution since the density of the reference substance is almost always water, and you can usually assume that the density of water is 1 gm/cc. So the volume occupied by 100 gm of solution is 100 / sp gr. Now you know the mass of substance per volume of solution in gm and cc. Convert everything to moles and liters and you have the molarity of the solution.
You should work through the calculation for yourself. It is a useful thing to be able to do. For concentrated nitric acid it should work out to 16 M, thereabouts.
Re: 0.5mol/lit HNO3Maybe this link will help.
http://www.chembuddy.com/?left=concentr ... centageq1 Also try this one. http://csmp.ucop.edu/downloads/csp/prep_percent.pdf
7 posts • Page 1 of 1
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