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 bo…

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asked Sep 27, 2020 24.2k views

1 Answer

Amaresh Jana · Answer 1 · 2020-10-02T16:27:33+0000

Answer:

14 g.

Step-by-step explanation:

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).