Question

How will you prepare 125 ml 0.03 M aqueous solution of KOH.Solve the given problem.

Answer 1

Steps:1) We know,

Molarity = No. of moles of solute / Volume of solution (in L)

2) Here, Malarity (KOH) ,M = 0.03 M

Volume of Solution ,V= 125mL

No. of Milli moles of KOH,n = MV

= 0.03*125 = 3.75 milli moles

3) Mass of KOH = n * MM(KOH)

=3.75 * 56 = 210mg = 0.21g

Therefore, 0.21g of KOH is added to 125mL of Water to produce 0.03M KOH solution.

Answer 2

Answer:

hey mate,, here is your answer..

Explanation:

let, M1 = 125ml

M2 = 0.03 M

V1 = 0.03 M

V2 = ?

therefore, M1 * V1 = M2 * V2

125 * 0.03 = 0.03 * V2

by solving you'll get,

V2 = 125 ml

hope it helps u frnd !