Suppose you were preparing 1.0 l of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of naocl. if all you had available was a jug of bleach that contained 0.82 m naocl, what volume of bleach would you need to add to the volumetric flask before you added enough water to reach the 1.0 l line?

To prepare 1.0 L of a bleaching solution that contains 0.21 mol of NaOCl, we need 0.26 L of a 0.82 M NaOCl solution.

We want to prepare 1.0 L (V₂) of a solution that contains 0.21 mol of NaOCl. The molar concentration (C₂) of the solution is:

We will start from a 0.82 M (C₁) NaOCl solution and dilute it. We can calculate the volume (V₁) of this concentrated solution using the dilution rule.

To prepare 1.0 L of a bleaching solution that contains 0.21 mol of NaOCl, we need 0.26 L of a 0.82 M NaOCl solution.

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0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

"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"

your answer would be “dry”