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

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.