Molar mass if calcium nitrate
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From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.
A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.
We know the formula for dilution type problems
M1 VI = M2 V 2
Where,
M, = initial molarity
V , = initial Volume
M2 = final molarity
V 2 = final Volume
Hene given -
M, = 3.00 M
VI = ?
M2 = 0.161M
V 2 = 175 ml
Accordingly ' MI V1 = M2 V 2
V1 =
V1= (0.161M*175ml)/ 3.00M
v1 = 9.391
The required volume of Stock solution is 9.391ml.
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