D STARTING AMOUNT X Your Aktiv Learning trial expires on 09/05/22 at 11:59 PM Question 28 of 29 What is the molality of lithium ions in a 0.302 m solution of Li,PO, assuming the compound dissociates completely? mol Li 6.022 x 10 0.001 35.0 mol Li,PO. Li,PO. 115.79 0.302 m Li m Li,PO. 3 0.906 1.21 0.101 g solvent kg solvent 1000 QU' Activate Now Submit

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pls help, I dont understand how this software wants me to enter the answer. its asking lfr the starting amount but its not taking it in the way Im putting it 
D
STARTING AMOUNT
X
Your Aktiv Learning trial expires on 09/05/22 at 11:59 PM
Question 28 of 29
What is the molality of lithium ions in a 0.302 m solution of Li,PO, assuming the
compound dissociates completely?
mol Li
6.022 x 10
0.001
35.0
mol Li,POL,PO.
115.79
0.302
m Li
m Li,PO
3
0.906
1.21
0.101
9 solvent kg solvent
1000
QU'
Activate Now
Submit
Transcribed Image Text:D STARTING AMOUNT X Your Aktiv Learning trial expires on 09/05/22 at 11:59 PM Question 28 of 29 What is the molality of lithium ions in a 0.302 m solution of Li,PO, assuming the compound dissociates completely? mol Li 6.022 x 10 0.001 35.0 mol Li,POL,PO. 115.79 0.302 m Li m Li,PO 3 0.906 1.21 0.101 9 solvent kg solvent 1000 QU' Activate Now Submit
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