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

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

(3) The ionized equations are,

Solution for Part 1 :

Formula used :

where,

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

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

= volume of NaCl stock solution

= volume of NaCl solution = 500 ml

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

     (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 at .

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

In 200 grams of water, the amount of sodium nitrate =

Now we have to calculate the mass of at .

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

In 200 grams of water, the amount of sodium nitrate =

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 precipitate will be, 52 grams

Solution for Part 3 :

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

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.