A chemist must dilute 61.9 mL of 548. nM aqueous sodium carbonate (Na2CO3) solution until the concentration falls to 484. nM . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.

The final volume will be "70.08 mL".

Explanation:

The given values are:

Molar mass,

M1 = 548 nM

or,

     =

M2 = 484 nM

or,

      =  

Volume,

V1 = 61.9 mL

V1 = ?

By using the expression, we get

⇒

or,

⇒            

By substituting the values, we get

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"