A35.0 g sample of a mixture of kcl and nacl required 199 ml of a 2.54 m agno3 solution to precipitate all the chloride as agcl. calculate the grams of kcl in the mixture.

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