In which of the following solutions would solid pbcl2 be expected to be the least soluble at
25°c?
a) 0.1 m cacl2
b) 0.1 m kno3
c) 0.1 m nacl
d) 0.1 m hcl
a) 0.1 m cacl2
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From the options the solution in which solid will be least soluble at  ( 25°C ) is ( A ); 0.1 M

Due to the phenomenon of common ion effect,   will be least soluble (100*) in a solution containing 0.1M of at 25°C .  

This is because when  undergoes ionization it gives away two chloride ions. Also when undergoes ionization it gives two chloride ions which is a Common ion. hence the common ion effect.  

The presence of common ions in solution decreases the solubility of a solute in a solvent . ( i.e. common ions between the solute and solvent reduces the solubility of the solute in the solvent significantly )

hence the solutions in which will be least soluble is 0.1 M

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A

Explanation:

As a result of common ion effect, lead II chloride will be 100 times less soluble in a solution that contains 0.1M chloride ions than it is in pure water.

CaCl2 ionizes to give two chloride ions. This implies about 10000 times less solubility in such a solution than it is in pure water hence the answer. Common ion effect means that if one of the ions in the solvent is common with one of the ions in the solute, the solubility of the solute in the solvent is decreased significantly because of the presence of a common ion.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%