# Mid Semester Ganjil Kimia Kelas Xii Ipa1 Tp. 2011-2012

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Soalan: 25 | Attempts: 821  Settings  Perhatian : 1. Mohon perhatikan waktu yang disediakan 2. Tidak meminta atau memberikan jawaban kepada teman 3. Jujurlah dengan apa yang ananda bisa

• 1.

### Diketahui beberapa larutan sebagai berikut : 1. 1 mol NaOH dalam 1 Liter air 2. 1 mol urea dalam 1 Liter larutan 3. 0,1 mol KOH dalam 100 gram air 4. 11,7 gran NaCl dalam 2000 gram air 5. 18 gram C6H12O6 dalam 100 gram etanol jika diketahui  Ar H=1, C=12, O=16, Na=23 dan Cl=35,5 maka larutan yang memiliki konsentrasi 1 molal adalah....

• A.

1,2 dan 3

• B.

1, 3 dan 5

• C.

2, 3 dan 5

• D.

2 dan 4

• E.

1 dan 3

C. 2, 3 dan 5
Explanation
The correct answer is 2, 3, and 5. This is because a solution is considered to be 1 molal if it contains 1 mole of solute dissolved in 1 kilogram of solvent. In option 2, there is 1 mole of urea dissolved in 1 liter of solution, which is approximately equal to 1 kilogram. In option 3, there is 0.1 mole of KOH dissolved in 100 grams of water, which is also approximately equal to 0.1 kilogram. In option 5, there is 18 grams of C6H12O6 dissolved in 100 grams of ethanol, which is approximately equal to 0.018 kilogram. Therefore, options 2, 3, and 5 meet the criteria for a 1 molal solution.

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• 2.

### Sebanyak 6 gram urea (Mr=60) dilarutkan dalam 90 gram air. Fraksi mol urea dalam larutan itu adalah....

• A.

0,0196

• B.

0,02

• C.

0,0625

• D.

0,0667

• E.

1,1

A. 0,0196
Explanation
The given question asks for the mole fraction of urea in the solution. To calculate the mole fraction, we need to find the moles of urea and water in the solution.

First, we calculate the moles of urea by dividing the given mass (6 grams) by the molar mass of urea (60 g/mol). This gives us 0.1 moles of urea.

Next, we calculate the moles of water by dividing the given mass (90 grams) by the molar mass of water (18 g/mol). This gives us 5 moles of water.

To find the mole fraction of urea, we divide the moles of urea by the total moles of solute (urea + water). This gives us 0.1 / (0.1 + 5) = 0.0196.

Therefore, the mole fraction of urea in the solution is 0.0196.

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• 3.

### Sebanyak x gram suatu zat nonelektrolit dengan massa molekul relatif M dilarutkan dalam 100 gram pelarut. penurunan titik beku larutan adalah T, maka harga penurunan titik beku molal (Kf) pelarut itu adalah....

• A.

(T . M)/10x

• B.

(100 . M)/10x

• C.

(100 . M) / (x . T)

• D.

(10x . T)/M

• E.

(1000 . M . x)/T

A. (T . M)/10x
Explanation
The correct answer is (T . M)/10x. This is because the formula for calculating the molal freezing point depression constant (Kf) is ΔTf = Kf * m, where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant, and m is the molality of the solution. In this case, the molality (m) is given by x grams of the solute divided by the mass of the solvent (100 grams). Therefore, the equation can be rearranged to solve for Kf, which gives (T . M)/10x.

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• 4.

### Larutan urea 0,1 molal membeku pada suhu -0,180C, pada saat yang sama larutan Na2SO4 0,1 molal membeku pada suhu -0,45 0C. Faktor vant Hoff (i) dari larutan Na2SO4 adalah.....

• A.

1

• B.

1,8

• C.

2,5

• D.

3

• E.

4,5

C. 2,5
Explanation
The freezing point depression is directly proportional to the molality and the van't Hoff factor (i) of a solution. In this case, both solutions have the same molality of 0.1 molal. However, the urea solution freezes at a higher temperature (-0.18°C) compared to the Na2SO4 solution (-0.45°C). This indicates that the Na2SO4 solution has a greater freezing point depression and therefore a higher van't Hoff factor (i). The only option that matches this is 2.5.

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• 5.

### Diantara peristiwa berikut ini yang tidak ada kaitannya dengan sifat koligatif larutan adalah....

• A.

Penggunaan glikol dalam cairan radiator mobil

• B.

Pencairan salju dijalan raya dengan manaburkan kristal garam

• C.

Penambahan tawas pada pengolahan air bersih

• D.

Membunuh lintah dengan menambahkan kristal garam

• E.

Naiknya air dari tanah ke puncak pohon

C. Penambahan tawas pada pengolahan air bersih
Explanation
The addition of alum (tawas) in the treatment of clean water is not related to the colligative properties of solutions. Colligative properties depend on the number of solute particles present in a solution, such as boiling point elevation, freezing point depression, and osmotic pressure. In the given options, the addition of glycol in the car radiator, the use of salt to melt snow, and the use of salt to kill leeches all involve the colligative properties of solutions. The movement of water from the ground to the top of a tree is also related to colligative properties, specifically osmotic pressure. Therefore, the correct answer is the addition of alum in the treatment of clean water.

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• 6.

### Kedalam 100 mL air dimasukkan 10 gram zat X non elektrolit, kemudian dibekukan dan diperoleh data sebagai berikut : Nomor      Waktu (menit)              Suhu Air (0C)              Suhu larutan (0C) 1                       5                                 4                                 0 2                       6                                 3                                 -1 3                       7                                 0                                 -2 4                       8                                 0                                 -3 5                       9                                 0                                 -3 6                      10                               -1                                 -3 Bila Kf air adalah 1,860C, maka Mr zat non elektrolit tersebut adalah......

• A.

34

• B.

60

• C.

62

• D.

90

• E.

139

C. 62
Explanation
The freezing point depression equation is ΔT = Kf * m, where ΔT is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solute. In this case, the freezing point depression is 3°C (0 - (-3)), and the Kf value for water is given as 1.86°C. By rearranging the equation, we can solve for the molality (m = ΔT / Kf). Substituting the values, we get m = 3 / 1.86 = 1.61 mol/kg. Since the solute is non-electrolyte, the molality is equal to the molar concentration (molarity). Therefore, the molar mass (Mr) of the solute is 1.61 g/mol. The closest option to this value is 62 g/mol.

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• 7.

### Kedalam 100 gram benzena dilarutkan 1,15 gram zat X nonelektrolit. Larutan ini membeku pada 4,940C. Jika diketahui titik beku benzena adalah 5,40C dan penurunan titik beku molalnya adalah 5,120C, maka massa molekul relatif zat X adalah....

• A.

(1,15 x 1000 x 4,49) / {100 x (5,4-5,12)}

• B.

(1,15 x 10 x 5,4) / {100 x (5,12-4,94)}

• C.

(1,15 x 1000 x 5,12) / {100 x (5,4-4,94)}

• D.

(1,15 x 5,12) / (5,4-4,94)

• E.

{100 X (5,4-4,94) / (1,15 X 1000 X 5,12)

C. (1,15 x 1000 x 5,12) / {100 x (5,4-4,94)}
Explanation
The given answer calculates the mass of the relative molecular weight of substance X by multiplying the mass of substance X (1.15 grams) by the freezing point depression constant (5.12°C) and dividing it by the difference in freezing points of the solvent (benzene) (5.4°C - 4.94°C) and the molality constant (100). This equation is derived from the equation for calculating the freezing point depression using molality.

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• 8.

### Larutan NaCl, KCl, MgSO4, K2SO4 dan H2SO4 masing-masing 0,1 m mempunyai faktor vant't Hoff berturut-turut 1,87; 1,86; 1,42; 2,46 dan 2,22. diantara larutan tersebut yang mempunyai titik didih tertinggi adalah.....

• A.

NaCl 0,1m

• B.

MgSO4 0,1m

• C.

H2SO4 0,1m

• D.

KCl 0,1m

• E.

K2SO4 0,1m

E. K2SO4 0,1m
Explanation
The factor vant't Hoff is a measure of the effect of a solute on the boiling point of a solvent. A higher factor vant't Hoff indicates a greater increase in boiling point. In this case, K2SO4 has the highest factor vant't Hoff of 2.46, which means it will cause the greatest increase in boiling point compared to the other solutes. Therefore, K2SO4 0.1m solution will have the highest boiling point among the given solutions.

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• 9.

### Suatu larutan elektrolit biner mendidih pada suhu 102,80C. Pada saat yang sama larutan urea dengan kemolalan yang sama mendidih pada 1020C. Derajat ionisasi elektrolit itu adalah....

• A.

0,2

• B.

0,4

• C.

0,6

• D.

0,8

• E.

1

B. 0,4
Explanation
The boiling point of a solution is affected by the presence of solute particles. In this case, the binary electrolyte solution has a boiling point of 102.8°C, while the urea solution with the same molarity boils at 102°C. This indicates that the binary electrolyte solution has a higher boiling point, suggesting that it has a greater degree of ionization. The answer of 0.4 indicates a moderate degree of ionization, which is consistent with the higher boiling point compared to the urea solution.

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• 10.

### Tingkat oksidasi wolfram dalam senyawa Na2W4O13.10H2O adalah.....

• A.

+4

• B.

+6

• C.

+8

• D.

+11

• E.

+12

B. +6
Explanation
The correct answer is +6. In the compound Na2W4O13.10H2O, the oxidation state of tungsten (W) is +6. This can be determined by considering the overall charge of the compound and the known oxidation states of the other elements. Sodium (Na) has an oxidation state of +1, and oxygen (O) has an oxidation state of -2. By assigning x as the oxidation state of tungsten, the equation becomes 2(+1) + 4x + 13(-2) + 10(0) = 0. Solving for x gives x = +6, indicating that tungsten is in the +6 oxidation state in this compound.

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• 11.

### Perhatikan reaksi redoks berikut : Cr2O7-2 + aH+ + bNO2-   → cCr+3 + dNO3- + eH2O harga koefisien reaksi diatas a, b, c, d, dan e berturut-turut adalah....

• A.

3,6,2,6,3

• B.

1,5,2,5,1

• C.

8,3,2,3,4

• D.

3,4,2,4,3

• E.

8,3,2,3,5

C. 8,3,2,3,4
Explanation
The balanced redox equation is as follows:
Cr2O7-2 + 14H+ + 6NO2- → 2Cr+3 + 6NO3- + 7H2O.

From the equation, we can see that the coefficients for H+ and NO2- are 14 and 6 respectively, which matches with the values given in the answer choice 8,3,2,3,4. The coefficients for Cr+3, NO3-, and H2O are 2, 6, and 7 respectively, which also matches with the values given in the answer choice 8,3,2,3,4.

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• 12.

### Cr(OH)4- + MnO4-  → MnO2 + Cr2O7-2 (belum setara, suasana basa). Jumlah mol Cr(OH)4- yang dapat dioksidasi oleh 1 mol MnO4-  adalah.....

• A.

3 mol

• B.

2 mol

• C.

1 mol

• D.

1/2 mol

• E.

1/3 mol

C. 1 mol
Explanation
In the given reaction, Cr(OH)4- is being oxidized by MnO4-. The balanced equation for the reaction is 1 Cr(OH)4- + 1 MnO4- -> 1 MnO2 + 1 Cr2O7-2. This means that 1 mol of Cr(OH)4- can be oxidized by 1 mol of MnO4-. Therefore, the correct answer is 1 mol.

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• 13.

### Pada reaksi redoks berikut :                 2FeCl2 + Cl2  → 2FeCl3 Reaksi setengah sel yang menunjukkan reaksi oksidasinya adalah....

• A.

Cl2 + 2e → 2Cl-

• B.

Fe+3 + e → Fe+2

• C.

Fe+2 + 2e → Fe

• D.

2Fe+2 + 3Cl2 + 4e → 2FeCl3

• E.

Fe+2 → Fe+3 + e

E. Fe+2 → Fe+3 + e
Explanation
The correct answer is Fe+2 → Fe+3 + e. This is the half-cell reaction that represents the oxidation of Fe+2 to Fe+3 by losing one electron.

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• 14.

### Pada reaksi :                 Cl2 + 2KOH → KClO + KCl + H2O Bilangan oksidasi klor berubah dari....

• A.

• B.

• C.

• D.

• E.

E. 0 menjadi -1 dan +1
Explanation
In the given reaction, Cl2 (chlorine) is being oxidized. The oxidation state of chlorine changes from 0 to -1 and +1. This means that in the reactants (Cl2), chlorine has an oxidation state of 0, and in the products (KClO and KCl), chlorine has oxidation states of -1 and +1 respectively. The change in oxidation state indicates that chlorine has gained one electron and lost one electron, resulting in the change from 0 to -1 and +1.

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• 15.

### Reaksi redoks :                 KMnO4 + K2C2O4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O Setengah reaksi oksidasi diatas adalah ....

• A.

MnO4- + 8H+ + 5e → Mn+2 + 4H2O

• B.

Mn+2 → Mn+7 + 5e

• C.

C2O4-2 → CO2 + 2e

• D.

4H+ + O2 + 4e → 2H2O

• E.

H2C2O4 + 2H+ + 2e → CO2 + 2H2O

C. C2O4-2 → CO2 + 2e
Explanation
The half-reaction given in the answer shows the oxidation of C2O4-2 to CO2. This means that C2O4-2 loses electrons and is oxidized to CO2. The coefficient of 2 in front of the electron indicates that 2 electrons are involved in the oxidation reaction. This is consistent with the overall reaction in which C2O4-2 is one of the reactants and CO2 is one of the products. Therefore, the given half-reaction is the correct representation of the oxidation process in the overall redox reaction.

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• 16.

### Diketahui potensial standar dari empat elektroda sebagai berikut :                 Cu+2 + 2e → Cu                  E0 = +0,34 v                 AgCl + e → Ag + Cl-          E0 = +0,22 v                 H+ + e → H2                        E0 = 0,00 v                 Zn+2 + 2e → Zn                 E0 = -0,76 v Potensial berikut yang merupakan potensial standar sel dari pasangan elektrode-elektrode diatas adalah....

• A.

0,42 volt

• B.

0,49 volt

• C.

0,54 volt

• D.

0,56 volt

• E.

0,98 volt

E. 0,98 volt
Explanation
The standard cell potential is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. In this case, the anode is Zn+2 + 2e -> Zn with a reduction potential of -0.76 V, and the cathode is Cu+2 + 2e -> Cu with a reduction potential of +0.34 V. Subtracting the anode potential from the cathode potential gives us 0.34 V - (-0.76 V) = 1.1 V. Therefore, the standard cell potential is 1.1 V, which is closest to the given answer of 0.98 V.

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• 17.

### Diketahui potensial standar beberapa sel volta sebagai berikut :                 A|A+2||B+2|B                     E0 = 3,50 V                 C|C+2||D+2|D                     E0 = 1,20 V                 C|C+2||B+2|B                      E0 = 1,75 V Maka potensial standar sel : A|A+2||D+2|D adalah....

• A.

2,95 V

• B.

6,45 V

• C.

4,05 V

• D.

0,35 V

• E.

4,7 V

A. 2,95 V
Explanation
The potential of a standard cell is determined by the difference in potential between the two half-cells involved. In this case, the potential of cell A|A+2||D+2|D can be calculated by subtracting the potential of cell A|A+2||B+2|B (3.50 V) from the potential of cell C|C+2||D+2|D (1.20 V). Therefore, the potential of cell A|A+2||D+2|D is 2.95 V.

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• 18.

### Jika diketahui :                 Zn + Cu+2 → Zn+2 + Cu   E0 = +1,10 Volt                 Sn+2 + 2e → Sn                  E0 = -0,14 Volt                 Cu+2 + 2e → Cu                 E0 = +0,34 Volt Maka potensial standar bagi reaksi Zn + Sn+2 →Zn+2 + Sn adalah....

• A.

1,44 V

• B.

1,24 V

• C.

0,96 V

• D.

0,76 V

• E.

0,62 V

E. 0,62 V
Explanation
The given question provides the standard reduction potentials for the reactions Zn + Cu+2 → Zn+2 + Cu, Sn+2 + 2e → Sn, and Cu+2 + 2e → Cu. To find the standard potential for the reaction Zn + Sn+2 → Zn+2 + Sn, we need to add the reduction potentials of the individual reactions. By adding the reduction potentials for the reactions Zn + Cu+2 → Zn+2 + Cu and Sn+2 + 2e → Sn, we get a total reduction potential of +1.10 V + (-0.14 V) = +0.96 V. Therefore, the correct answer is 0.96 V.

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• 19.

### Logam dapat dilindungi dari korosi dengan cara berikut , kecuali ....

• A.

Diolesi minyak

• B.

Diolesi larutan garam dapur

• C.

Dilapisi timah

• D.

Dilapisi cat

• E.

Disepuh

B. Diolesi larutan garam dapur
Explanation
The correct answer is "diolesi larutan garam dapur" because saltwater solutions can actually accelerate the corrosion process on metals. When metal is exposed to saltwater, it creates an electrolyte solution that enhances the flow of electrons, leading to increased corrosion. Therefore, applying saltwater solution to metal surfaces would not protect them from corrosion.

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• 20.

### Dari data potensial elektrode standar berikut :                 Cu+2 + 2e → Cu                 E0 = +0,34 V                 Ag+ + e → Ag                      E0 = +0,80 V Maka reaksi Cu + 2Ag+ → Cu+2 + 2Ag memiliki potensial sel .....

• A.

0,06 Volt

• B.

0,46 Volt

• C.

0,56 Volt

• D.

1,14 Volt

• E.

1,26 Volt

B. 0,46 Volt
Explanation
The potential of a cell can be determined by subtracting the reduction potential of the anode from the reduction potential of the cathode. In this case, the reduction potential of Cu+2 to Cu is +0.34 V and the reduction potential of Ag+ to Ag is +0.80 V. Subtracting the reduction potential of the anode (Cu+2 to Cu) from the reduction potential of the cathode (Ag+ to Ag), we get +0.80 V - +0.34 V = +0.46 V. Therefore, the potential of the cell is 0.46 Volt.

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• 21.

### Diketahui data potensial elektrode sebagai berikut :                 Ag+ + e → Ag                      E0 = +0,80 Volt                 Mg+2 + 2e → Mg               E0 = -2,34 Volt                 In+3 + 3e → In                   E0 = -0,34 Volt                 Mn+2 + 2e → Mn               E0 = -1,20 Volt Pasangan sel dibawah ini yang memberikan perbedaan potensial 1,14 volt adalah.....

• A.

Ag|Ag+||In+3|In

• B.

Mn|Mn+2||Ag+|Ag

• C.

Mn|Mn+2||Mg+2|Mg

• D.

In|In+3||Ag+|Ag

• E.

Mg|Mg+2||Ag+|Ag

D. In|In+3||Ag+|Ag
Explanation
The given answer, In|In+3||Ag+|Ag, is the correct answer because it is the only combination that has a potential difference of 1.14 volts. The reduction potential for In+3 to In is -0.34 volts, and the reduction potential for Ag+ to Ag is +0.80 volts. When we subtract the reduction potential of In+3 to In from the reduction potential of Ag+ to Ag, we get a potential difference of 1.14 volts. Therefore, this combination is the only one that satisfies the given condition.

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• 22.

### Diketahui data potensial elektrode sebagai berikut :                 Cu+2 + 2e → Cu                 E0 = +0,34 Volt                 Fe+3 + e → Fe+2                 E0 = +0,77 Volt                 Pb+2 + 2e → Pb                 E0 = -0,33 Volt                 Cu+2 + e →Cu+                   E0 = +0,15 Volt Dari data tersebut maka reaksi yang tidak dapat berlangsung spontan adalah....

• A.

F+3 + Pb → Fe+2 + Pb+2

• B.

Cu+2 + Pb → Cu+ + Pb+2

• C.

Pb+2 + Cu → Pb + Cu+2

• D.

Fe+3 + Cu → Fe+2 + Cu+2

• E.

Fe+3 + Cu+ → Fe+2 + Cu+2

C. Pb+2 + Cu → Pb + Cu+2
Explanation
The reaction Pb+2 + Cu → Pb + Cu+2 cannot occur spontaneously because the standard electrode potential (E0) for this reaction is negative (-0.33 Volt). In spontaneous reactions, the E0 value should be positive, indicating a tendency for the reaction to occur. Therefore, the given reaction cannot proceed spontaneously.

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• 23.

### Data dari reaksi setengah sel dengan E0 sebagai berikut :                 Pb+2 / Pb             = -0,13 Volt                 Sn+2 / Sn             = -0,14 Volt                 Mg+2/ Mg              = -2,37 Volt                 Cu+2/Cu               = +0,34 Volt                 Ni+2/Ni                  = -0,25 Volt                 Fe+2/Fe                 = -0,44 Volt Logam yang dapat mencegah korosi pipa besi yang ditanam didalam tanah adalah....

• A.

Mg

• B.

Cu

• C.

Sn

• D.

Ni

• E.

Pb

A. Mg
Explanation
Based on the given data, the metal with the highest reduction potential (most positive value) is Mg with a reduction potential of -2.37 Volt. This means that Mg has a higher tendency to be reduced and therefore acts as a sacrificial anode, preventing the corrosion of the iron pipe. Therefore, Mg is the metal that can prevent the corrosion of the iron pipe when it is buried in the ground.

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• 24.

• A.

Pt dan C

• B.

Zn dan C

• C.

Pb dan PbO2

• D.

Zn dan Cu

• E.

Cu dan PbO2

C. Pb dan PbO2
Explanation
The correct answer is Pb dan PbO2. This is because lead (Pb) and lead dioxide (PbO2) are commonly used as the positive and negative electrodes in lead-acid batteries. PbO2 acts as the positive electrode (cathode) while Pb acts as the negative electrode (anode) during the discharge process. This combination allows for the reversible conversion between lead sulfate and lead dioxide, which is essential for the functioning of the battery.

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• 25.

### Jika diketahui :                 E0 Pb+2/Pb = -0,13 Volt                 E0 Fe+2/Fe = -0,44 Volt Maka sel volta yang disusun dengan menggunakan elektrode Pb dan Fe memiliki beda potensial sebesar.....

• A.

-0,31 Volt

• B.

+0,31 Volt

• C.

-0,54 Volt

• D.

+0,54 Volt

• E.

+0,57 Volt Back to top