Find percentage mass of koh present in 0.12 molal aqueous solution in it
Answers
Answered by
1
Answer:
1.288 g mL
−1
100 g solution contains 30 g KOH
Therefore ,
d
100
mL solution contains
56
30
mol KOH , d=density in g/m
2
Molarity =6.90=
56
30
×
100
d
×1000
d=1.288 g/mL
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