Answer :
(1) The volume must dilute to make 500 mL of a 0.50M NaCl solution is, 0.125 L
(2) The amount of [tex]NaNO_3[/tex] precipitate will be, 52 grams
(3) The ionized equations are,
[tex]NH_4Cl(aq)\rightarrow NH_4^+(aq)+Cl^-(aq)\\\\Cu(NO_3)_2(aq)\rightarrow Cu^{2+}(aq)+2NO_3^-(aq)\\\\CH_3COOH(aq)\rightarrow CH_3COO^-(aq)+H^+(aq)\\\\HgCl_2(aq)\rightarrow Hg^{2+}(aq)+2Cl^-(aq)[/tex]
Solution for Part 1 :
Formula used : [tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = concentration of NaCl stock solution = 2 M = 2 mole/L
[tex]M_2[/tex] = concentration of NaCl solution = 0.50 M = 0.50 mole/L
[tex]V_1[/tex] = volume of NaCl stock solution
[tex]V_2[/tex] = volume of NaCl solution = 500 ml
Now put all the given values in the above formula, we get the volume of NaCl stock solution.
[tex](2mole/L)\times V_1=(0.50mole/L)\times (500ml)[/tex]
[tex]V_1=125ml=0.125L[/tex] (1 L = 1000 ml)
Therefore, the volume must dilute to make 500 mL of a 0.50M NaCl solution is, 0.125 L
Solution for Part 2 :
First we have to calculate the mass of [tex]NaNO_3[/tex] at [tex]50^oC[/tex].
In 100 grams of water, the amount of sodium nitrate = 114 g
In 200 grams of water, the amount of sodium nitrate = [tex]\frac{114}{100}\times 200=228g[/tex]
Now we have to calculate the mass of [tex]NaNO_3[/tex] at [tex]20^oC[/tex].
In 100 grams of water, the amount of sodium nitrate = 88 g
In 200 grams of water, the amount of sodium nitrate = [tex]\frac{88}{100}\times 200=176g[/tex]
Now we have to calculate the amount of sodium nitrate precipitated.
The amount of sodium nitrate precipitated = 228 - 176 = 52 g
Therefore, the amount of [tex]NaNO_3[/tex] precipitate will be, 52 grams
Solution for Part 3 :
When the substance dissolved in water then they disassociate into respective ions.
[tex]NH_4Cl(aq)\rightarrow NH_4^+(aq)+Cl^-(aq)\\\\Cu(NO_3)_2(aq)\rightarrow Cu^{2+}(aq)+2NO_3^-(aq)\\\\CH_3COOH(aq)\rightarrow CH_3COO^-(aq)+H^+(aq)\\\\HgCl_2(aq)\rightarrow Hg^{2+}(aq)+2Cl^-(aq)[/tex]
