The quantity of required dry solute is <u>4.1g KCl for 107 g of 0.535m soln.</u>
<u>Explanation:</u>
<u>
Given:</u>
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.
The answer is b.
i hope i helped!
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