Answer:
18 KJ
Step-by-step explanation:
Data Given:
mass of Lead (m) = 21 g
Heat taken for vaporization (Q) = ?
Solution:
This problem is related to phase change and latent heat of vaporization.
Latent heat of vaporization is the amount of heat taken to convert one mole of substance at its boiling point to its vapor.
So, Latent heat of vaporization of lead has a constant value
Latent heat of vaporization of lead = 177.7 KJ/mol
Formula used
Q = m x Lv. . . . . (1)
where
Lv = specific latent heat of vaporization
here the value for latent heat of vaporization is for mole so instead of mass we will use moles in formula.
So,
Q = no. of mol x Lv. . . . . (2)
first find no. of moles for 21 g of lead
no. of moles = mass in grams / molar mass . . . . . . (3)
molar mass of lead (Pb) = 207 g/mol
put values in equation 3
no. of moles = 21g / 207 g/mol
no. of moles = 0.101 mol
so,
number of moles of lead (Pb) = 0.101 mol
Put values in the eq.2
Q = 0.101 mol x 177.7 KJ/mol
Q = 18 kJ
So, 18KJ of heat is taken to vaporize 21 g of lead (Pb)