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Chemistry
Tutor: None Selected Time limit: 1 Day

Mar 13th, 2015

a) Molar mass of Magnesium Nitrate, Mg(NO3) 2 = 24.3050 + (14.0067 + 15.9994 * 3) *2

  = 148.3148 g/mol

So, Magnesium Nitrate, Mg(NO3) 2 = 0.523/148.3148 =0.0035 mol

Molarity (M)= Amount (mol) of solute / Volume (L) of solution=0.0035/(235/1000) = 0.015 M  (Answer)

b)  Molar mass of LiClO4.3H2O = 6.941+ 35.453+15.999*4+3*(1.007*2+15.999) 

  = 160.429 g/mol

So, LiClO4.3H2O = 20.2/160.429 = 0.126 mol

Molarity = 0.126/0.230 = 0.547 M (Answer)

c) As, we know, V1XS1 = V2XS2

=> 40*2.64 = 0.390*1000*S2

So, S2 = 0.271 (Answer)


Mar 14th, 2015

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