Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

Answer 1

The quantity of required dry solute is 4.1g KCl for 107 g of 0.535m soln.

Explanation:

Given:

MM of KCl is 74.55

107g of 0.535m KCl

0.535 molal = 0.535 mols KCl/kg solvent.

0.535 mols KCl = 74.55 x 0.535

                          = 39.88 g KCl.

Therefore a 0.535 m soln consists of 39.88 g KCl + 1000 g H2O for a total of 1039.88 g solution.

To get 107 g of solution

39.88 x (107/1038.88) = 4.1g KCl for 107 g of 0.535m soln.  

The quantity of required dry solute is 4.1g KCl for 107 g of 0.535m soln.