posted by kelseyyy on .
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.