blueislife1436 blueislife1436
  • 30-08-2017
  • Chemistry
contestada

How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?

Respuesta :

LeChatNoir
LeChatNoir LeChatNoir
  • 30-08-2017
C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol

mKI: 39+127 = 166 g/mol

1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI

:)
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