the solubility of lead(II) chloride, PbCl2, in water at 20 degrees Celcius is 1.00 g PbCl2/100 g of water. if you stirred 7.50 g PbCl2 in 400 g of water at 20 degrees Celcius, what mass of lead (II) chloride would remain undissolved?
Solubility PbCl2 is 1.0 g in 100 g H2O.
So solubility in 400 g water is 4 x that or 1.0 g PbCl2 x (400/100) = 4.0 g.
Subtract from 7.5 g to find the amount undissolved.
To determine the mass of lead(II) chloride that would remain undissolved, we can use the concept of solubility and compare it with the amount of lead(II) chloride you stirred into the water.
The solubility of lead(II) chloride in water tells us that at 20 degrees Celsius, 1.00 g of PbCl2 will dissolve in 100 g of water. This means that the solution is saturated when these proportions are reached.
Now, let's calculate the maximum amount of lead(II) chloride that can dissolve in 400 g of water using the solubility information provided:
1.00 g PbCl2/100 g H2O = x g PbCl2/400 g H2O
Cross-multiplying, we get:
x g PbCl2 = (1.00 g PbCl2/100 g H2O) * 400 g H2O
x g PbCl2 = 4.00 g PbCl2
So, in 400 g of water, you can dissolve a maximum of 4.00 g of lead(II) chloride.
However, you stirred in 7.50 g of PbCl2, which is more than the maximum solubility. Therefore, the excess lead(II) chloride will remain undissolved.
To determine the mass of lead(II) chloride that remains undissolved, subtract the maximum solubility from the amount you stirred in:
Mass of PbCl2 remaining undissolved = Amount stirred in - Maximum solubility
Mass of PbCl2 remaining undissolved = 7.50 g - 4.00 g
Mass of PbCl2 remaining undissolved = 3.50 g
Therefore, 3.50 g of lead(II) chloride would remain undissolved.