Molarity MðÞ¼
Number of moles of solute
Litre of solution
¼
Weight in gram 1000
Molecular weightvolume mlðÞ
For example, the molarity of a 1 litre solution containing 4 g of NaOH will be
0.1 M, i.e.,
Molarity¼
4 1000
40 1000
¼ 0 :1M
If we know the molarity of solution to be prepared, then the amount (weight) of
salt that needs to be dissolved for preparation of per liter solution can be calculated.
For example, to prepare 1 litre solution of 0.2 M NaOH,
Molecular weight of NaOH¼40, so
0 : 2 ¼
Weight in gram 1000
40 1000
Weight in gram¼0.2 40 ¼8 g (i.e., 8 g of NaOH is dissolved in 1 litrefinal
volume of distilled water).
Alternatively, if we directly multiply required molarity with molecular weight, it
will give weight of salt in gram per liter of solution.
For example, to calculate amount of salt to prepare 0.2 M NaOH,
0 : 2 40 ¼8g=L
Note If required molarity is in mM, then salt amount will be in mg/liter, and if
molarity is inμM, then amount will be inμg/liter.
However, the molarity of conc. solution can be expressed as
M¼
%purityspecific gravity 1000
Molecular weight 100
6.2 Normal Solutions.................................
One normal solution is the solution which contains 1 g equivalent weight of solute
per liter of solution.
22 6 Preparation of General Laboratory Solutions and Buffers