25.0 mL of a 6.0 M HNO_3 stock solution is diluted using water to 100 mL. How ma

Question

25.0 mL of a 6.0 M HNO_3 stock solution is diluted using water to 100 mL. How many moles of HNO_3 are present In the dilute solution? 0.5 moles 1.5 moles 5.0 moles 0.15 moles How many mL of a 14.0 M NH_3, stock solution are needed to prepare 200 mL of a 4.20 M dilute NH_3 solution? 667 mL 0.060 mL 840 mL 60.0 mL The four following concentrated solutions are each diluted with water to form 200 mL of a dilute solution. Which solution once diluted to 200 mL will have the largest concentration? 80.0 mL of a 0.2 M NaOH solution 100.0 mL of a 0.1 M NaOH solution 10.0 mL of a 0.5 M NaOH solution 20.0 mL of a 0.4 M NaOH solution

Explanation / Answer

Part A:

moles= molarity * Volume in L

We have 6 M, 25 ml HNO3 solution,

then, we will convert 25 ml to L, as, 25/1000= 0.025 L,

By putting the values to the equation,

moles= 6*0.025,

moles= 0.15 moles HNO3,

part B :

By using formula Mi*Vi=MfVf,

where, i= initial , f=final , M= molar concentration , v= volume,

we have MiVi=MfVf,

14*Vi =4.20*0.20 L

Vi=MfVf/Mi,

Vi=4.20*0.20 L/14,

Vi= 0.84/14,

Vi=0.06 L (0.06*1000= 60 mL) ,

PART C:

Moles = molarity * volume in L,

MOLES= 0.2 *0.08 L=0.0016 moles,

moles= 0.1*0.10=0.01 moles,

moles= 0.50*0.01=0.005 moles,

moles= 0.4*0.02=0.008 moles,

after diluting the solution to 200 ml moles will remains same,

so, 100 ml 0.1 M solution will have highest molar concentration after dilution. (0.01 moles)

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