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?

Answer 1

Answer :

(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.

$(2mole/L)\times V_1=(0.50mole/L)\times (500ml)$

$V_1=125ml=0.125L$      (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 = $\frac{114}{100}\times 200=228g$

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 = $\frac{88}{100}\times 200=176g$

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.

$NH_4Cl(aq)\rightarrow NH_4^+(aq)+Cl^-(aq)\\\\Cu(NO_3)_2(aq)\rightarrow Cu^{2+}(aq)+2NO_3^-(aq)\\\\CH_3COOH(aq)\rightarrow CH_3COO^-(aq)+H^+(aq)\\\\HgCl_2(aq)\rightarrow Hg^{2+}(aq)+2Cl^-(aq)$