Chemistry, asked by mosley, 1 year ago

how many moles of NaOH are contained in 27 ml 0.15 M NaOH.????​

Answers

Answered by Anonymous
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 rinayjainsl
0

Answer:

0.00405 moles of NaOH are present in the given solution.

Explanation:

Given data is,

Volume of NaOH solution is V=27ml=0.027L

Molarity of NaOH solution is M=0.15M

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-

M=\frac{n}{V(in\:lit)}

Substituting the values of volume and molarity of solution in the above relation,we get

0.15=\frac{n}{0.027} = > n=0.15\times0.027=0.00405mol

Therefore,0.00405 moles of NaOH are present in the given solution.

#SPJ2

Similar questions