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

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.