Mass of the sample (NaCl + KCl ) = 35.0 g
Concentration of AgNO3 = 2.54 M
Volume of AgNo3 = 199 ml = 0.199 L
Moles of AgNO3 = 2.54 moleL-1 * 0.199 L = 0.5055 moles
Since all the Ag in AgCl has to come from AgNO3, then
Moles of AgNO3 = moles of AgCl = 0.5055
Now, since all the Cl- comes from NaCl and KCl, we can write:
# Moles NaCl + # moles KCl = 0.5055
Therefore, # moles NaCl = 0.5055 - moles KCl--------(1)
It is given that
Mass NaCl + Mass KCl = 35.0 g
i.e.
(58 g/mole)NaCl * #Moles NaCl + (75 g/mol)KCl * #Moles KCl = 35.0 g---(2)
from eq(1) we have:
58 (0.5055-moles KCl) + 75 *moles KCl = 35.0
Therefore,
moles KCl = 5.681/133 = 0.0427 moles
Now, molar mass of KCl = 75 g/mol
Thus, the mass of KCl in the mixture = 75 g/mole * 0.0427 moles = 3.203 g
