Respuesta :
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.
When The quantity of required dry solute is 4.1g KCl for 107 g of 0.535m soln.
Solvent
When MM of KCl is 74.55
Then 107g of 0.535m KCl
After that 0.535 molal = 0.535 mols KCl/kg solvent.
Then 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.
Then To get 107 g of solution
So that 39.88 x (107/1038.88) = 4.1g KCl for 107 g of 0.535m soln.
Thus, The quantity of required dry solute is 4.1g KCl for 107 g of 0.535m soln.
Find out more information about solvent here:
https://brainly.com/question/23946616
