Do 85 g wody dodano 25 g tlenku sodu, otrzymano roztwór o stężeniu
A. 25,00 %.
B. 29,41 %.
C. 32,26 %.
D. 4,0 mol/dm3
A. 25,00 %.
B. 29,41 %.
C. 32,26 %.
D. 4,0 mol/dm3
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mNaOH=40u
Na₂O + H₂O ---> 2NaOH
62g Na₂O ------ 80g NaOH
25g Na₂O ------ xg NaOH
x = 32,26g NaOH
ms=32,26g
mr=100g
Cp=ms*100%/mr
Cp=32,26*100/100
Cp = 32,26%