The following questions all refer to an aqueous 8.41 M potassium hydroxide stock
ID: 912396 • Letter: T
Question
The following questions all refer to an aqueous 8.41 M potassium hydroxide stock solution. If 29.6 mL of the stock solution are diluted to 477 mL, what is the molarity (in mol/L) of the new solution? 5.22 Times 10^-1 M If 10.1 mL of the stock solution are diluted to obtain a 0.750 M potassium hydroxide solution, what is the volume (in mL) of the new solution? 1.13 Times 10^2 mL Determine the volume (in mL) of the original stock solution that must be used to prepare 277 mL of a 0.692 M potassium hydroxide solution. 2.28 times 10^1 mL Determine the volume of water (in mL) that was added to 21.2 mL of the stock solution to prepare a 0.152 M potassium hydroxide solution.Explanation / Answer
formulae M1V1=M2V2
1 .M1=MOLARITY OF STOCK SOLUTION(8.41) , M2=MOLARITY OF NEW SOLUTION(?)
V1=VOUME OF STOCK SOLUTION(29.6) ,V2= VOLUME OF NEW SOLUTION(477)
M1V1=M2V2 ===>8.41*29.6= M2*477
M2 =5.22*10-1M
2 .M1=8.41,M2=0.750,V1=10.1,V2=?
Acc to M1V1=M2V2
8.41*10.1=0.750*V2
V2=1.13*102ml
3. M1=8.41, V1=?, M2=0.692, V2=277
Acc to.M1V1=M2V2
8.41*V1=0.692*277
V1=2.28*101ML
4. M1=8.41 , V1=21.2 , M2=0.152 ,V2=?
Acc to M1V1=M2V2
8.41*21.2=0.152*V2
V2=1173ml
Water to be added=1173-21.2
=1151.8 ml
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