10) An aqueous solution withdensity 0.988 g mL–1 at 20 oC is prepared by dissolv

Question

10) An aqueous solution withdensity 0.988 g mL–1 at 20 oC is prepared by dissolving 12.8mL CH3CH2CH2OH (d = 0.803 g mL–1) inenough water to produce 75.0 mL of solution. What is thepercent CH3CH2CH2OH expressed as

(a) percent byvolume:

(b) percent bymass:

(c) w/v percent(mass/volume %):

Explanation / Answer

a) 12.8 mL/ 75 mL = 0.170666667 => 17.1 % b) 12.8 mL * ( 0.803 g / mL) = 10.2784 grams Ch3CH3CH3OH mass of solution = 75 mL * (0.988 g/mL) = 74.1 grams totla % mass = 10.2784 g / 74.1 g = 0.138709852 => 13.9 % c) 10.2784 g / 75 mL = 0.137045333 => 13.7 g/mL %

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