Question

1.You have a 2.0M NaCl stock solution available. What is the volume you must dilute to make 500 mL of a 0.50M NaCl solution?

2. How many grams of NaNO3 will precipitate if a saturated solution of NaNO3 in 200 g of water at 50°C is cooled to 20oC?
Assume the following solubility values for NaNO3: 114.0g/100g H2O at 50oC; 88.0g/100g H2O at 20oC

3. Write equations to show how these substances ionize or dissociate in water.
NH4Cl
Cu(NO3)2
HC2H3O2
HgCl2

Answer 1

(1) The volume must dilute to make 500 mL of a 0.50M NaCl solution is, 0.125 L

(2) The amount of
`NaNO_3` precipitate will be, 52 grams

(3) The ionized equations are,

`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)`

Solution for Part 1 :

Formula used :
`M_1V_1=M_2V_2`

where,

`M_1` = concentration of NaCl stock solution = 2 M = 2 mole/L

`M_2` = concentration of NaCl solution = 0.50 M = 0.50 mole/L

`V_1` = volume of NaCl stock solution

`V_2` = volume of NaCl solution = 500 ml

Now put all the given values in the above formula, we get the volume of NaCl stock solution.

`(2mole/L)* V_1=(0.50mole/L)* (500ml)`

`V_1=125ml=0.125L` (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
`NaNO_3` at
`50^oC`.

In 100 grams of water, the amount of sodium nitrate = 114 g

In 200 grams of water, the amount of sodium nitrate =
`(114)/(100)* 200=228g`

Now we have to calculate the mass of
`NaNO_3` at
`20^oC`.

In 100 grams of water, the amount of sodium nitrate = 88 g

In 200 grams of water, the amount of sodium nitrate =
`(88)/(100)* 200=176g`

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
`NaNO_3` precipitate will be, 52 grams

Solution for Part 3 :

When the substance dissolved in water then they disassociate into respective ions.

`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)`