Answer to Question #90751 in General Chemistry for Adediran Jeremiah

Question

An alloy of mass 588g and volume 100cm3 is made of iron of density 8.0g/cm3 and aluminium of density 2.7g/cm3. Calculate the proportion (i) by volume, (ii) by mass of constituents of the alloy

Answer 1

1 cm³ of alloy would weigh 588/100 = 5.88 g

1 cm³ of iron would weigh = 8.0 g

1 cm³ of aluminum would weigh = 2.7 g

Let proportion of iron be x and that of aluminum would be 1-x

Thus,

8.0 * x + 2.7 * (1-x) = 5.88

8x+2.7-2.7x=5.88

X= 0.6 i.e 60% by mass is iron and 40% by mass is aluminum.

1g. of alloy would have volume = 100/588 = 0.17 cm³

1 g. of iron would have volume = 1/8 = 0.125 cm³

1 g. of aluminum would have volume = 1/(2.7) = 0.37 cm³

Let y be the volume proportion of iron and 1-y for aluminum.

Thus, 0.125*y+0.37*(1-y) = 0.17

0.125y+0.37-0.37y=0.17

Y=0.816

i.e 81.6% of volume is occupied by iron and rest 18.4% by aluminum.