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How to prepare 0.1M 100mL AgNo3 (MW = 169.8731 g/mol)
So, what we have is,
M=0.1M
V=100mL/0.1L
MW=169.8731g/mol
Weight = ?
Formula:-
mol = M x V = Weight / MW
thus weight = M x V x MW
= 0.1 x 0.1 x 169.8731 (mol/L x g/mol x L)
= 1.69g
Measure 1.69g of AgNO3 and put inside a 100mL volumetric flask. top up with distilled H2O until 100mL. Siap!
To prepare a 0.1 N solution of silver nitrate, dissolve 1.7 grams of silver nitrate (AgNO3) in enough distilled water to make 1 liter of solution. This concentration corresponds to 0.1 moles of silver nitrate per liter, which is equivalent to 0.1 Normal (N) solution.
The easiest way is to get a litre volumetric flask. Find out the molarity or the %V/V. If you have molarity, it is a simple exercise to find out how much you need by dilution factor.. If you have %V/V, you need to get the density. This gives you the weight per volume. Then you can get the moles per volume via no. moles is mass/molecular weight. You can now work out the molarity of the stock by bringing the volume to 1000. Now you do the same as before, i.e. dilution factor.
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