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## Solutions Exam2 and Problem Solutions

**Solutions Exam2 and Problem Solutions**

**1.** Solubility vs. temperature graph of X solid is given below. Using this graph decide, which ones of the following statements are true;

**I.** When X is dissolved in water, temperature of water decreases.

**II.** 200 g solution under 35 ^{0}C, using 60 g X is saturated solution.

**III.** When 50 g saturated solution at 35 ^{0}C is cooled to 15 ^{0}C, 5 g X crystallizes.

**Solution:**

**I.** As you can see from the graph, solubility of X in water increases with increasing temperature. Thus, solubility of X in water in endothermic. When X dissolves in closed container, it absorbs heat from water and as a result temperature of water decreases. I is true.

**II.** At 35 ^{0}C;

100 g water dissolves 30 g X

200 g water dissolves ? g X

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=60 g X can be dissolved

Since amount of X is 60 g in 200 g solution, it is saturated solution. II is true.

**III**. 100 water can dissolve 30 g X at 35 ^{0}C and 20 g X at 15 ^{0}C. When solution prepared under 35 ^{0}C is cooled to 15 ^{0}C;

30-20=10 X is crystallized.

In 100 g water 10 g X is crystallized

In 50 g water ? g X is crystallized

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=5 g X is crystallized in 50 g water

III is also true.

**2.** If solubility of sugar in water is endothermic, which ones of the following statements increase both solubility of sugar and solubility rate?

**I.** Cooling solution

**II.** Using granulated sugar instead cube sugar

**III.** Mixing the solution

**IV. **Increasing amount of sugar

**V.** Increasing temperature of solution

**Solution:**

II, III and IV do not affect solubility. In endothermic solutions increasing temperature increases solubility of that matter. Moreover, increasing temperature also increases solubility rate. Thus, V increase both solubility and solubility rate of sugar in water.

**3. **We add 700 mL water at same temperature to 0,2 molar 300 mL NaCl solution. Find final molarity of this solution.

**Solution:**

M_{1}=0,2 molar

V_{1}=300 mL

V_{2}=700+300=1000 mL

We use dilution formula;

M_{1}.V_{1}=M_{2}.V_{2}

0,2.300=M_{2}.1000

M_{2}=0,06 molar

**4. **9,8 g H_{2}SO_{4} is dissolved in water and 200 mL solution is prepared. Find normality of solution.(H_{2}SO_{4}=98)

**Solution:**

There is a relation between normality and molarity;

N=M.Equivalent

n_{H2SO4}=9,8/98=0,1mol H_{2}SO_{4}

M=n/V=0,1/0,2=0,5 molar

V=200 mL=0,2 L

N=M.Equivalent (Where equivalent is 2 since H_{2}SO_{4} gives 2 H^{+} ion to solution)

N=0,5.2=1N

**5.** 0,4 mol MgCl_{2} and 0,6 mol AlCl_{3} are dissolved in water and 250 mL solution is prepared. Find molar concentration of [Cl^{-}] in this solution.

**Solution:**

We write ionization reactions of both salts and find number of moles of ions;

MgCl_{2}(s) → Mg^{+2}(aq) + 2Cl^{-}(aq)

0,4mol 0,4mol 0,8mol

AlCl_{3}(s) → Al^{+3}(aq) + 3Cl^{-}(aq)

0,6mol 0,6mol 1,8mol

Mole of Cl- ion = 0,8 + 1,8 =2,6mol

Volume of Solution=250mL=0,25L

[Cl^{-}]=n_{Cl-}/Vsol.=2,6/0,25=10,4molar

The Original Author: Mrs. Şerife (Erden) SARICA

Tags: Solutions | Concentration with Examples | Solutions Exams and Problem Solutions | Solutions Exam1 and Problem Solutions | Solutions Exam3 and Problem Solutions |

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