Question

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.

Answer 1

Answer:

The final volume will be "70.08 mL".

Explanation:

The given values are:

Molar mass,

M1 = 548 nM

or,

     = $5.48\times 10^{-7} \ M$

M2 = 484 nM

or,

      = $4.84\times 10^{-7} \ M$  

Volume,

V1 = 61.9 mL

V1 = ?

By using the expression, we get

⇒ $M1\times V1=M2\times V2$

or,

⇒            $V2=\frac{M1\times V1}{M2}$

By substituting the values, we get

                    $=\frac{5.48\times 10^{-7}\times 61.9}{4.84\times 10^{-7}}$

                    $=\frac{339.212}{4.84}$

                    $=70.08 \ mL$