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.

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

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