Answer / kang chuen tat

(a) Let X = specific gravity of the 60 % mixture.
By interpolation for a straight line with constant m as gradient, then (75 - 40) / (0.877 - 0.952) = (60 - 40) / (X - 0.952). Then 35 / (-0.075) = -466.67 = 20 / (X - 0.952). X = -0.0429 0.952 = 0.9091.
(b) Let V = the required volume of the 40 % mixture.
Mass of ethanol in 75 % mixture Mass of ethanol in 40 % mixture = Mass of ethanol in 60 % mixture.
Then, 300 (0.877) (0.75) V (0.952) (0.4) = (V 300) (0.9091) (0.6).
Then 197.325 0.3808 V = 0.5455 V 163.638. V = 204.53 gallons.

The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS 61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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