Molarity of a 250 mL solution containing 5.5 g of potassium bromide c) volumee i

Question

Molarity of a 250 mL solution containing 5.5 g of potassium bromide c) volumee in liters of 3.46 M potassium hydroxide that contains 5.35g solute. D) Moles of solute in 5.0 L of 0.850 M sodium sulfate e) Moles of sulfate in 5.0 L of 0.850 M sodium sulfate f) number of CI^- ions in 52 L f 2.3 M copper (II) chloride. 8. A 0.7 M solution of vitamin B12 must be diluted 1:5 in order to obtain the absorbance. How would you prepare 100 mL of the diluted solution? b). What is the concentration of the diluted solution? 9. A stock solution of protein, 8 mg/mL is to be used to prepare a 50 g/mL solution. How would you prepare 10 mL of this diluted solution? 10. how would you use serial dilution to prepare 10 mL of a 1.0 times 10^-6M solution of sodium chloride (NaCI) from a 0.1 M stock solution?

Explanation / Answer

b)

MW KBr = 119.002

mol = mass/MW = 5.5/119.002 = 0.0462177 mol

M = mol/L = 0.0462177 / 0.25 = 0.18487M of KBr

c)

m = 5.35 g of solute KOH

MW = 56.1056

mol = mass/MW = 5.35/56.1056 = 0.09535

then

M = mol/L

L = mol/M = 0.09535/3.46 = 0.02755780346L = 27.5 ml

d)

V = 5 L

M = 0.85

mol = m*V = 5*.085 = 0.425 moles of solute

e)

moles of sulfate!

0.425 moles of Na2SO4

then

0.425  mol of 2Na+ and SO4-2

answer must be 0.425 moles of SO4-2

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