A mixture of toluene and benzene is analyzed and found to contain 47percentage t

Question

A mixture of toluene and benzene is analyzed and found to contain 47percentage toluene by mass. Calculate the liters of this mixture required to give 350 mol benzene. The s g of benzene and toluene are 0.879 and 0.866 respectively. A mixture contains 5 mol pentane and 10 mol hexane. Calculate the mass fraction of the components in the mixture and the average molecular weight At 25 degree C, an aqueous solution containing 35.0 wt percentage H_2SO_4 has a specific gravity of 1.2563. A quantity of the 35 percentage solution is needed that contains 195.5 kg of H2So_4. Calculate the required volume (L) of the solution using the given specific gravity. Estimate the percentage error that would have resulted if pure component specific gravities of H_2SO_4(S.G.=1 8255) and water had been used for the calculation instead of the given specific gravity of the mixture.

Explanation / Answer

1. note: unit of s.g. is not given. here it is assumed that unit should be gm/ml

mass of tolune is 47% by mass then mass of benzene will be 53% or for simplification in 100 gm of a mixture weight of tolune is 47 gm and benzene is 53 gm. so, the volume of this 100 gm solution will be = (53/0.879) ml + (47/0.866) = 114.57 ml

mole of benzene = (53/78) = 0.679 [ molecular weight of benzene 78 gm]

hence, 0.679 mole benzene obtained from 114.57 ml mixture.

so, 350 mole will obtained from = (350*114.57/ 0.679) = 59056.7 ml = 59.0567 liter of mixture.

2. molecular weight of pentane is 72 gm /mol

weight of 5 mol will be = 72 * 5 = 360 gm

molecular weight of pentane is 86 gm /mol

weight of 10 mol will be = 86 * 10 = 860 gm

total mass = 360 + 860 =1220 gm

mass fraction of pentane = (360/1220) = 0.295

mass fraction of hexane = (860/1220) = 0.7049

average molecular weight = 1220/2 = 610 gm

3. a. weight of H2SO4 is 35gm and H2O is 65 gm. the volume of solution will be = 100 / 1.2563=79.59 ml

35 gm of H2SO4 obtained from 79.59 ml of solution

so, 195.5 kg will obtained = (79.59*195.5*1000)/(35*1000) = 444.56 L

b. volume of H2SO4 will be = 35/1.8255 = 19.17 ml and volume of water is 65 ml. total volume will be = 65 +19.17=84.17 ml

35 gm of H2SO4 obtained from 84.17 ml of solution

so, 195.5 kg will obtained = (84.17*195.5*1000)/(35*1000) = 470.14 L

% of error will be = [(470.14 - 444.56) /444.56]*100 = 5.75 %

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