how many moles of NaOH are contained in 27 ml 0.15 M NaOH.????
Answers
Answered by
16
Molarity, M = 0.15 M
Volume, V, in L = 27 ml = 0.027 L
Moles of solute NaOH = M × V = 0.15 × 0.027 = 0.0041moles
Substituting values, moles of solute NaOH = 0.15 × 0.027 = 0.0041moles
Answered by
0
Answer:
0.00405 moles of NaOH are present in the given solution.
Explanation:
Given data is,
Volume of NaOH solution is
Molarity of NaOH solution is
And we are required to find the number of moles of NaOH contained in this solution.For that we use molarity relation as mentioned below-
Substituting the values of volume and molarity of solution in the above relation,we get
Therefore,0.00405 moles of NaOH are present in the given solution.
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