Problem:
How many ml of 1% potassium chloride would be needed to precipitate all of the silver in a 0.5 g ore sample that contains 1.5 parts per thousand silver? Allow for a 50% excess of the chloride solution.

Solution:
To find the volume of 1% potassium chloride solution required to precipitate all of the silver in the ore sample, we need to consider the following steps:

Step 1: Calculate the mass of silver in the ore sample:
Given that the ore sample contains 1.5 parts per thousand silver and the mass of the ore sample is 0.5 g, we can calculate the mass of silver in the ore sample using the following equation:

Mass of silver = (1.5/1000) * mass of ore sample
= (1.5/1000) * 0.5 g
= 0.00075 g

Step 2: Convert the mass of silver to moles:
To convert the mass of silver to moles, we need to divide the mass by the molar mass of silver. The molar mass of silver (Ag) is 107.87 g/mol.

Number of moles of silver = mass of silver / molar mass of silver
= 0.00075 g / 107.87 g/mol
= 6.952 x 10^-6 mol

Step 3: Calculate the volume of 1% potassium chloride solution needed:
To calculate the volume of 1% potassium chloride solution needed, we need to determine the stoichiometry of the reaction between silver and potassium chloride. The balanced equation for the reaction is:

2 AgNO3 + KCl → AgCl + 2 KNO3

From the balanced equation, we can see that 2 moles of silver react with 1 mole of potassium chloride to form 1 mole of silver chloride. Therefore, the molar ratio between silver and potassium chloride is 2:1.

Since we have 6.952 x 10^-6 mol of silver, we need half the moles of potassium chloride for complete precipitation. Therefore,

Number of moles of potassium chloride = 6.952 x 10^-6 mol / 2
= 3.476 x 10^-6 mol

Now, we can calculate the volume of 1% potassium chloride solution needed using the following equation:

Volume of potassium chloride solution = (Number of moles of potassium chloride / concentration of potassium chloride) * 1000
= (3.476 x 10^-6 mol / 0.01 mol/L) * 1000
= 347.6 mL

Step 4: Include a 50% excess of the chloride solution:
To account for a 50% excess of the chloride solution, we need to multiply the calculated volume by a factor of 1.5.

Volume of potassium chloride solution (including 50% excess) = 347.6 mL * 1.5
= 521.4 mL

Therefore, approximately 521.4 mL of 1% potassium chloride solution would be needed to precipitate all of the silver in the ore sample, allowing for a 50% excess of the chloride solution.