Question
. How many grams of potassium chloride, KCl, are needed to prepare 0.630 L of a 2.50 M solution of potassium chloride? (4.0 points) (Points :4)
1.575 M
117 g
2.54 m
23.19 g
1.575 M
117 g
2.54 m
23.19 g
Answers
vishal
solve by
molarity =no. of moles divide by volume in litres.......
so let the mass be in x grms...
then :
2.50*0.630=x grms
so,
x grms of KCl =1.575 grams is the answer
molarity =no. of moles divide by volume in litres.......
so let the mass be in x grms...
then :
2.50*0.630=x grms
so,
x grms of KCl =1.575 grams is the answer