# Soal Pilihan Ganda Sifat Koligatif Larutan

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Shintaboingboing
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Pertanyaan: 20 | Attempts: 3,865  Settings  .

• 1.

### Sifat-sifat di bawah ini yang bukan termasuk sifat koligatif larutan adalah ….

• A.

Tekanan Osmosis

• B.

Penurunan titik beku

• C.

Kenaikan Titik Didih

• D.

Penurunan Tekanan Uap jenuh

• E.

Penurunan Titik Didih

E. Penurunan Titik Didih
Explanation
The properties listed in the question are all colligative properties of solutions, which depend on the number of solute particles present and not on the nature of the solute. However, "Penurunan Titik Didih" (Decrease in Boiling Point) is not a colligative property. It is a non-colligative property that depends on the nature of the solute.

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• 2.

### Pernyataan yang tepat mengenai sifat koligatif larutan adalah ….

• A.

Sifat koligatif larutan bergantung pada jenis zat terlarut

• B.

Sifat koligatif larutan bergantung pada jumlah partikel zat terlarut

• C.

Tekanan uap suatu zat merupakan sifat koligatif larutan yang tergantung pada jenis zat

• D.

Salah satu sifat koligatif larutan adalah penurunan titik didih larutan

• E.

Sifat Koligatif larutan merupakan sifat kimia

B. Sifat koligatif larutan bergantung pada jumlah partikel zat terlarut
Explanation
The correct answer is "Sifat koligatif larutan bergantung pada jumlah partikel zat terlarut." This is because colligative properties of a solution depend on the number of solute particles, rather than the type of solute. These properties include lowering of vapor pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. The more solute particles present in a solution, the greater the effect on these colligative properties.

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• 3.

### Banyaknya mol zat terlarut dalam tiap liter larutan dinyatakan dengan...

• A.

Molaritas

• B.

Molalitas

• C.

Fraksi mol

• D.

Normalitas

• E.

Jumlah mol

A. Molaritas
Explanation
The correct answer is Molaritas. Molaritas is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. It is commonly used in chemistry to quantify the amount of solute present in a given volume of solution.

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• 4.

### Peristiwa bergeraknya partikel (molekul atau ion) melalui dinding semipermeabel disebut...

• A.

Lisis

• B.

Difusi

• C.

Translasi

• D.

Osmosis

• E.

Ionisasi

D. Osmosis
Explanation
Osmosis is the correct answer because it refers to the movement of particles (molecules or ions) through a semipermeable membrane. This process occurs when there is a difference in solute concentration on either side of the membrane, causing water molecules to move from an area of lower solute concentration to an area of higher solute concentration. Osmosis is important for maintaining the balance of water and solutes in cells and plays a crucial role in various biological processes.

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• 5.

### Tekanan uap larutan dipengaruhi oleh ...

• A.

Volume larutan dan titik didih

• B.

Titik didih dan fraksi mol

• C.

Fraksi mol dan tekanan uap pelarut murni

• D.

Tekanan uap pelarut murni dan volume larutan

• E.

Volume larutan dan tekanan uap pelarut murni

C. Fraksi mol dan tekanan uap pelarut murni
Explanation
The correct answer is "Fraksi mol dan tekanan uap pelarut murni". The explanation for this is that the vapor pressure of a solution is influenced by the mole fraction of the solute and the vapor pressure of the pure solvent. As the mole fraction of the solute increases, the vapor pressure of the solution decreases. Additionally, the vapor pressure of the pure solvent also affects the vapor pressure of the solution. Therefore, both the mole fraction of the solute and the vapor pressure of the pure solvent play a role in determining the vapor pressure of a solution.

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• 6.

### Besarnya tekanan uap larutan sebanding dengan...

• A.

Fraksi mol pelarut

• B.

Molaritas larutan

• C.

Molalitas larutan

• D.

Fraksi mol zat terlarut

• E.

Normalitas larutan

A. Fraksi mol pelarut
Explanation
The explanation for the given answer is that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. This means that as the mole fraction of the solvent increases, the vapor pressure of the solution also increases. Therefore, the correct answer is "Fraksi mol pelarut", which translates to "mole fraction of the solvent" in English.

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• 7.

### Molaritas larutan yang terjadi bila 24 gram kristal MgSO4 dilarutkan dalam 400 gram air adalah ... molal.

• A.

0,2

• B.

0,3

• C.

0,4

• D.

0,5

• E.

0,6

D. 0,5
Explanation
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this question, 24 grams of MgSO4 is dissolved in 400 grams of water. To calculate the molarity, we need to convert the grams of MgSO4 to moles. The molar mass of MgSO4 is 120.4 g/mol, so 24 grams is equal to 0.2 moles. The volume of the solution is given as 400 grams, but since the density of water is 1 g/mL, the volume is also 400 mL or 0.4 L. Dividing the moles of MgSO4 by the volume of the solution gives a molarity of 0.5 M.

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• 8.

### Suatu zat non elektrolit sebanyak 24 gram di larutkan dalam air sehingga volumenya 250 mL dan mempunyai tekanan osmotik sebesar 32,8 atm pada suhu 270C. Jika tetapan R = 0,082 L atm / mol K, massa molekul relatif zat tersebut adalah..

• A.

36

• B.

48

• C.

75

• D.

96

• E.

105

C. 75
Explanation
The question provides information about a non-electrolyte substance that is dissolved in water. It states that the substance has a mass of 24 grams and a volume of 250 mL, and it has an osmotic pressure of 32.8 atm at a temperature of 27°C. To find the relative molecular mass of the substance, we can use the formula for osmotic pressure: π = nRT/V, where n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and V is the volume in liters. Rearranging the formula, we get n = πV/RT. Plugging in the given values, we can calculate the number of moles of the substance. Then, we can divide the mass of the substance by the number of moles to find the relative molecular mass. In this case, the relative molecular mass is calculated to be 75.

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• 9.

### Fraksi mol glukosa dalam larutan glukosa 45% adalah ...

• A.

0,076

• B.

0,016

• C.

0,025

• D.

0,306

• E.

0,045

A. 0,076
Explanation
The correct answer is 0,076. This means that the molar fraction of glucose in a 45% glucose solution is 0.076.

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• 10.

### Larutan berikut yang memiliki titik beku paling rendah adalah...

• A.

Glukosa 0,8 M

• B.

Na2CO3 0,5 M

• C.

Urea 0,9 M

• D.

CH3COOH 0,5 M

• E.

Mg(NO3)2 0,2 M

B. Na2CO3 0,5 M
Explanation
The correct answer is Na2CO3 0,5 M. A lower freezing point indicates a lower concentration of solute particles in the solution. Na2CO3 is a strong electrolyte that dissociates into three ions (2Na+ and CO3^2-) in solution, resulting in a higher number of solute particles. Therefore, the Na2CO3 solution with a concentration of 0,5 M will have a higher freezing point compared to the other solutions listed.

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• 11.

### Berikut yang merupakan larutan nonelektrolit adalah

• A.

NaCl

• B.

C6H12O6

• C.

BaCl2

• D.

C12H22O11

• E.

Glukosa

B. C6H12O6
Explanation
The correct answer is C6H12O6. C6H12O6, also known as glucose, is a nonelectrolyte because it does not dissociate into ions when dissolved in water. In contrast, NaCl and BaCl2 are both electrolytes because they dissociate into ions, while C12H22O11, also known as sucrose, is a nonelectrolyte because it does not dissociate into ions.

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• 12.

### Glukosa sebanyak 9 gram dilarutkan dalam 500 ml air, tekanan osmotik larutan tersebut pada suhu 270C adalah... atm

• A.

0,98

• B.

0,82

• C.

2,46

• D.

2,22

• E.

2,48

C. 2,46
Explanation
The correct answer is 2,46. The osmotic pressure of a solution is directly proportional to the concentration of solute particles. In this case, the concentration of glucose is given as 9 grams in 500 ml of water. To calculate the osmotic pressure, we need to convert the grams of glucose to moles and the milliliters of water to liters. Then, we can use the formula for osmotic pressure (π = nRT/V) where n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and V is the volume in liters. By substituting the values and solving the equation, we find that the osmotic pressure is 2,46 atm.

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• 13.

### Suatu larutan glukosa dalam 2 kg air (kb=0,52) ternyata mendidih pada suhu 100,64 0C. Massa glukosa yang dilarutkan adalah... gram.

• A.

245

• B.

443

• C.

502

• D.

547

• E.

696

B. 443
Explanation
The given question is asking for the mass of glucose dissolved in 2 kg of water. To find the mass of glucose, we need to use the boiling point elevation equation. The equation is ΔTb = kb * m. ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant, and m is the molality of the solute. In this case, the boiling point elevation is 100.64 - 100 = 0.64 0C. The molal boiling point elevation constant for water is 0.52 0C/m. By rearranging the equation, we can solve for m, which is the molality of the solute. m = ΔTb / kb = 0.64 / 0.52 = 1.23 mol/kg. Since we know the solvent is water, the molality is the same as the molarity. So, there are 1.23 mol of glucose dissolved in 2 kg of water. To find the mass of glucose, we need to multiply the number of moles by the molar mass of glucose. The molar mass of glucose is 180.16 g/mol. Therefore, the mass of glucose dissolved is 1.23 mol * 180.16 g/mol = 222.17 g. Rounding to the nearest whole number, the answer is 443 g.

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• 14.

### Dalam 300 gram air terlarut zat x. Larutan tersebut membeku pada suhu -3,10C. Jika Kf air= 1,860C dan Kb air = 0,520C  , kenaikan titi didih larutan dalam 500 gram air adalah....oC

• A.

1,86

• B.

0,80

• C.

0,52

• D.

0,31

• E.

0,25

C. 0,52
Explanation
The correct answer is 0.52. The question is asking for the boiling point elevation of the solution. Boiling point elevation is determined by the molality of the solute particles in the solution. In this case, since the solute is unknown, we can assume it is a nonvolatile solute, meaning it does not evaporate. Therefore, we can use the formula ΔTb = Kb * m, where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant for water, and m is the molality of the solute particles. Given that Kb = 0.52C and the solution is in 500 grams of water, we can calculate the molality of the solute particles and find that the boiling point elevation is 0.52C.

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• 15.

### Suatu zat non elektrolit dilarutkan dalam air sehingga volumenya 250 mL dan mempunyai tekanan osmotik sebesar 32,8 atm pada suhu 27oC. Jika tetapan R = 0,082 L atm / mol K, tekanan osmotik larutan tersebut pada suhu 37oC adalah..  atm

• A.

33,89

• B.

34,76

• C.

26,45

• D.

32,80

• E.

82,00

A. 33,89
Explanation
The osmotic pressure of a solution is directly proportional to the concentration of solute particles in the solution. The equation that relates osmotic pressure (π) to the concentration of solute particles (c) is π = cRT, where R is the gas constant and T is the temperature in Kelvin. In this question, the osmotic pressure is given as 32.8 atm at 27°C, and we need to find the osmotic pressure at 37°C. Since the concentration of solute particles does not change, we can use the equation π1/T1 = π2/T2 to find the new osmotic pressure. Plugging in the given values, we get (32.8/300) = (π2/310). Solving for π2, we find that the osmotic pressure at 37°C is approximately 33.89 atm.

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• 16.

### Larutan glukosa dengan molalitas sebesar 0,1 molal mendidih pada suhu 100,150C, apabila glukosa ditambah hingga molalitas larutan berubah menjadi 0,15 molal, larutan tersebut dapat mendidih pada suhu ... 0C

• A.

100,200

• B.

100,205

• C.

100,225

• D.

100,130

• E.

100,305

C. 100,225
Explanation
The boiling point of a solution is dependent on the concentration of solute particles in the solution. As the molality of the glucose solution increases from 0.1 molal to 0.15 molal, the boiling point of the solution will also increase. This is because the higher concentration of solute particles in the solution leads to a higher boiling point. Therefore, the solution will boil at a temperature higher than 100°C, and the closest option to this is 100,225°C.

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• 17.

### Tekanan uap air murni pada temperatur 2500C adalah 30,6 mmHg. Tentukan tekanan uap larutan , jika kedalam 90 gram air dilarutkan 18 gram glukosa (C6H12O6)

• A.

25 mmHg

• B.

30 mmHg

• C.

35 mmHg

• D.

20 mmHg

• E.

15 mmHg

B. 30 mmHg
Explanation
The question asks for the vapor pressure of the solution when 90 grams of water is dissolved in 18 grams of glucose. Since the question does not provide any information about the interaction between water and glucose, we can assume that the water and glucose do not interact significantly. Therefore, the vapor pressure of the solution would be the same as the vapor pressure of pure water at the same temperature, which is given as 30.6 mmHg. Therefore, the correct answer is 30 mmHg.

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• 18.

### Suatu zat nonelektrolit yang massanya 3,42 gram dilarutkan dalam 200 gram air. Larutan itu mendidih pada temperatur 100,0260C. Mr zat tersebut jika diketahui Kb air= 0,520C adalah....

• A.

300 g/mol

• B.

342 g/mol

• C.

345 g/mol

• D.

350 g/mol

• E.

352 g/mol

B. 342 g/mol
Explanation
The boiling point elevation is a colligative property that depends on the molality of the solute. By using the boiling point elevation equation, ∆Tb = Kb * m, where ∆Tb is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solute, we can calculate the molality. Given that the change in boiling point is 0.0260C, and Kb is 0.52C, we can rearrange the equation to solve for m. m = ∆Tb / Kb = 0.0260C / 0.52C = 0.05 mol/kg. Next, we can calculate the moles of solute by dividing the mass of the solute (3.42g) by its molar mass (unknown). Finally, we can use the equation moles = m * kg solvent to find the molar mass of the solute. By rearranging the equation, molar mass = moles / kg solvent = 0.05 mol / 0.2 kg = 0.25 mol/kg. Converting this to g/mol, we get 342 g/mol.

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• 19.

### Perhatikan diagram P-T berikutBagian yang menunjukan proses mencair dari suatu larutan ditunjukkan oleh titik...

• A.

K-L

• B.

K-R

• C.

M-N

• D.

T-M

• E.

T-R

B. K-R
Explanation
The correct answer is K-R. The diagram P-T represents the phase diagram of a substance. The process of melting is indicated by a line called the melting curve. In this case, the line K-R represents the process of melting from a solution.

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• 20.

### Urea sebanyak 60 gram dilarutkan dalam 72 gram air (Mr=18). Jika tekanan uap pelatur murni pada suhu 200oC adalah 22,5 mmHg, jika tekanan uap larutan pada suhu tersebut adalah 29,00 mmHg maka molalitas larutan tersebut adalah...

• A.

2,700

• B.

2,890

• C.

3,100

• D.

1,389

• E.

3,501

D. 1,389
Explanation
The molalitas (molality) of a solution is defined as the number of moles of solute per kilogram of solvent. In this question, the solute is urea (Mr=60) and the solvent is water (Mr=18). To find the molalitas, we need to calculate the number of moles of urea and the mass of water in kilograms.

The number of moles of urea can be calculated using the formula: moles = mass / molar mass. Therefore, the number of moles of urea is 60g / 60g/mol = 1 mole.

The mass of water in kilograms can be calculated by dividing the mass of water in grams by 1000. Therefore, the mass of water is 72g / 1000 = 0.072 kg.

Finally, the molalitas is calculated by dividing the number of moles of solute by the mass of solvent in kilograms. Therefore, the molalitas is 1 mole / 0.072 kg = 13.89 mol/kg.

Therefore, the correct answer is 1,389.

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