STARTING AMOUNT X What volume in ADD FACTOR x( ) 0.559 559 mol Nal L of a 0.724 M Nal solution contains 0.405 mol of Nal ? 0.724 22.99 g Nal 6.022 x 1023 126.90 M Nal ANSWER 0.293 mL 0.405 148.89 L 1.79 RESET 3 g Nal/mol 1 +

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Chapter15: Solutions
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Question
STARTING AMOUNT
X
What volume in
ADD FACTOR
x( )
0.559
559
mol Nal
L of a 0.724 M Nal solution contains 0.405 mol of Nal ?
0.724
22.99
Question 12 of 24
g Nal
6.022 × 1023
126.90
M Nal
=
ANSWER
0.293
mL
0.405
148.89
L
1.79
RESET
3
g Nal/mol
1
Submit
+
Transcribed Image Text:STARTING AMOUNT X What volume in ADD FACTOR x( ) 0.559 559 mol Nal L of a 0.724 M Nal solution contains 0.405 mol of Nal ? 0.724 22.99 Question 12 of 24 g Nal 6.022 × 1023 126.90 M Nal = ANSWER 0.293 mL 0.405 148.89 L 1.79 RESET 3 g Nal/mol 1 Submit +
Expert Solution
Step 1

Molarity of NaI = 0.724 M = 0.724 moles/L (Molarity, M = moles/L)

Number of moles of NaI = 0.405 moles

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