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You mix 3 solutions: 0.5 L of 10 mM Tris-HCl 1.5 mL of 1 M Tris-HCl...

You mix 3 solutions: 0.5 L of 10 mM Tris-HCl 1.5 mL of 1 M Tris-HCl 200 mL of NaCl at 1.2%(w/v) (MW: 58.44) What is the final concentration of NaCl, in mM?

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Answer #1

1L=1000 mL

0.5 L= 0.5\times1000=500 mL

final volume= 500+200+1.5=701.5 mL

701.5 mL=701.5/1000=0.7015 L

for NaCl

M1=1.2%

M2=?

V1=200 mL

V2= 701.5 mL

M1V1=M2V2

M2= M1V1/V2

= (1.2\times200)/701.5

=0.342 %

(mass/volume)\times100=0.342%

(mass/701.5)=0.342%/100

mass= (0.342/100)\times701.5

= 2.4 g

number of moles= mass in gram/gram molecular mass

= 2.4/58.44

= 0.041

molarity= number of moles/volume of the solution in litres

= 0.041/0.7015

= 0.0585 M

1 M= 1000 mM

so the concentration of NaCl in mM=58.5 mM

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