A magnesium-lead alloy of mass 5.5 kg consists of a solid α phase that has a composition that is just slightly below the solubility limit at 200C (390F).(a) What mass of lead is in the alloy?(b) If the alloy is heated to 350C (660F), how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase?

We have to calculate (for the magnesium-lead alloy) the mass of lead in 5.5 kg of the solid α phase at 200°C just below the solubility limit. The solubility limit for the α phase at 200°C is about 5 wt% Pb.

The mass of lead in the alloy = (0.05)*(5.5 kg) = 0.28 kg

b) If the alloy is heated to 350°C, how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase?

At 350°C, the solubility limit of the a phase increases to approximately 25 wt% Pb.

C(lead) = ((mass fo lead in alloy) + (mass lead)) / ((magnesium-lead alloy mass) + (mass lead))

0.25 = (0.28 + m(Pb)) / ( 5.5 + m(Pb))

0.25 * ( 5.5 + m(Pb)) = (0.28 + m(Pb))

1.375 + 0.25 m(Pb) = 0.28 + m(Pb)

1.095 = 0.75 m(Pb)

m(Pb) = 1.095 / 0.75

m(Pb) = 1.46

There can be dissolved 1.46 kg more of lead.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T13:43:02+02:00

The mass of lead and how much more lead may be dissolved are mathematically given as

mPb= 0.28 kg

m(Pb) = 1.46kg

What is the mass of lead and how much more lead may be dissolved?

Question Parameter(s):

A magnesium-lead alloy of mass 5.5 kg

The solubility limit at 200C (390F)

If the alloy is heated to 350C (660F),

a)

Generally, the equation for the mass of lead is mathematically given as

at 200

200 °C -> 5 wt% Pb

Therefore

mPb= (0.05)*(5.5 kg)

mPb= 0.28 kg

b)

[tex]C= \frac{(mass fo lead in alloy) + (mass lead0}{ ((magnesium-lead alloy mass) + (mass lead)) }[/tex]

0.25 = (0.28 + m(Pb)) / ( 5.5 + m(Pb))

1.375 + 0.25 m(Pb) = 0.28 + m(Pb)

m(Pb) = 1.46

In conclusion

m(Pb) = 1.46kg