Question

A beaker containing 80 grams of lead(ii) nitrate, pb(no3)2, in 100 grams of water has a temperature of 30 ºc. approximately how many grams of the salt are undissolved, on the bottom of the beaker?

Answer 1

Answer:

14 g.

Explanation:

• From the figure attached:

the solubility of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC is (66 g).

When beaker containing 80 grams of lead(II) nitrate, Pb(NO₃)₂, in 100 grams of water has a temperature of 30ºC.

∴ The grams of the salt are undissolved, on the bottom of the beaker are (14 g).