Quant Chemistry (MCQ Test)

  • Grade 11th,
  • Grade 12th
  • ACS
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| By Shuvro Chakravorty
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Shuvro Chakravorty
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Quizzes Created: 1 | Total Attempts: 227
| Attempts: 227 | Questions: 30 | Updated: Apr 20, 2025
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1. āύāĻŋāĻšā§‡āϰ āϕ⧋āύ āĻ†ā§Ÿāύ āϜāĻžāϰāĻ• āĻ“ āĻŦāĻŋāϜāĻžāϰāĻ• āωāϭ⧟ āĻšāĻŋāϏ⧇āĻŦ⧇ āĻ•āĻžāϜ āĻ•āϰ⧇?

Explanation

Fe2+ can function as both a reducing agent and an oxidizing agent. As a reducing agent, it can donate electrons to other substances and get oxidized in the process. As an oxidizing agent, it can accept electrons from other substances and get reduced in the process. This ability to undergo both reduction and oxidation reactions makes Fe2+ versatile and capable of acting as both a reducing agent and an oxidizing agent.

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About This Quiz
Quant Chemistry (MCQ Test) - Quiz

This 'Quant Chemistry' quiz assesses knowledge in quantitative chemical analysis, focusing on reactions, stoichiometry, and electron transfer in various compounds. It is designed for learners to apply theoretical concepts practically, enhancing their understanding of chemical properties and reactions.

2.

What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2. āĻ—ā§āϝāĻžāϏ āĻ•ā§āϰ⧋āĻŽāĻžāĻŸā§‹āĻ—ā§āϰāĻžāĻĢāĻŋāϤ⧇ āĻŦā§āϝāĻŦāĻšāĻžāϰāϝ⧋āĻ—ā§āϝ āĻŦāĻžāĻšāĻ• āĻ—ā§āϝāĻžāϏ āϕ⧋āύāϟāĻŋ?

Explanation

In gas chromatography, a carrier gas is used to carry the sample through the column. The carrier gas should be inert and not react with the sample components. Among the given options, N2 (nitrogen gas) is the most suitable choice as it is inert and does not react with most compounds. O2 (oxygen gas) can react with certain compounds, Cl2 (chlorine gas) is highly reactive and can cause unwanted reactions, and H2 (hydrogen gas) is flammable and poses a safety risk. Therefore, N2 is the correct answer.

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3. Ca(OCl)Cl āĻ…āϪ⧁āϤ⧇ Cl āĻāϰ āϜāĻžāϰāĻŖ āĻ…āĻŦāĻ¸ā§āĻĨāĻž-

Explanation

The oxidation state of Cl in Ca(OCl)Cl is +1 and -1. In the compound, the oxidation state of Ca is +2, and since the overall charge of the compound is 0, the sum of the oxidation states of all the elements must be 0. Therefore, the oxidation state of Cl must be such that when added to the oxidation state of Ca (+2), it gives a sum of 0. Since there are two Cl atoms, one Cl atom must have an oxidation state of +1 and the other Cl atom must have an oxidation state of -1 in order to satisfy this condition.

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4. āĻĻā§āϰāĻŦāϪ⧇āϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻžāϰ āϕ⧋āύ āĻāĻ•āĻ• āϤāĻžāĻĒ āύāĻŋāĻ°ā§āĻ­āϰāĻļā§€āϞ?

Explanation

The correct answer is "মোলারিটি" because it is the unit of concentration that measures the amount of a substance in a given volume of a solution. It is dependent on temperature as it represents the number of moles of solute per liter of solution. The other options, "মোল ভগ্নাংশ" and "মোলালিটি", are not units of temperature-dependent concentration.

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5. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻĒā§āϰāĻžāχāĻŽāĻžāϰ⧀ āĻ¸ā§āĻŸā§āϝāĻžāĻ¨ā§āĻĄāĻžāĻ°ā§āĻĄ āĻĒāĻĻāĻžāĻ°ā§āĻĨ?

Explanation

K2Cr2O7 is the correct answer because it is the only compound in the given options that is a primary standard substance. Primary standard substances are highly pure and stable compounds that can be used to accurately determine the concentration of a solution in titration experiments. Na2S2O3, NaOH, and FeSO4 are not primary standard substances.

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6. 5g FeSO4 āϕ⧇ āĻ…āĻŽā§āĻ˛ā§€ā§Ÿ āĻŽāĻžāĻ§ā§āϝāĻŽā§‡ āϜāĻžāϰāĻŋāϤ āĻ•āϰāϤ⧇ āĻ•āϤ āĻ—ā§āϰāĻžāĻŽ KMnO4 āĻĒā§āĻ°ā§Ÿā§‹āϜāύ?

Explanation

not-available-via-ai

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7. 5% āϭ⧇āϜāĻžāϞāϝ⧁āĻ•ā§āϤ 100 g āϚ⧁āύāĻžāĻĒāĻžāĻĨāϰ āϤ⧀āĻŦā§āϰ āϤāĻžāĻĒ⧇ āĻŦāĻŋāϝ⧋āϜāĻŋāϤ āĻ•āϰāϞ⧇ āĻ•āϤ āĻ—ā§āϰāĻžāĻŽ CaO āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ?

Explanation

The question asks for the amount of CaO that can be obtained when 100g of limestone containing 5% impurities is heated at high temperature. Since limestone is mainly composed of CaCO3, the impurities will not contribute to the formation of CaO. Therefore, only 95% of the 100g of limestone will be converted to CaO. 95% of 100g is equal to 95g.

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8. 0.1 M āϏāĻŽāĻ†ā§ŸāϤāύ⧇āϰ NaOH āĻ“ H2SO4 āĻāϰ āĻĻā§āϰāĻŦāϪ⧇āϰ āĻĒā§āϰāĻ•ā§ƒāϤāĻŋ āĻ•āĻŋāϰ⧂āĻĒ āĻšāĻŦ⧇?

Explanation

The nature of the solution formed by mixing 0.1 M NaOH and H2SO4 will be acidic because H2SO4 is a strong acid and NaOH is a strong base. When they react, they will form H2O and Na2SO4, which is a salt. The presence of H2SO4 in the solution will make it acidic.

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9. āĻŽāĻŋāĻĨāĻžāχāϞ āĻ…āϰ⧇āĻžā§āĻœā§‡āϰ āĻŦāĻ°ā§āĻŖ āĻĒāϰāĻŋāĻŦāĻ°ā§āϤāύ⧇āϰ pH āĻĒāϰāĻŋāϏāϰ-

Explanation

The correct answer is 3-5. This pH range indicates that the color change of the methyl orange indicator occurs when the solution is acidic. Methyl orange is an acid-base indicator that changes color from red to yellow when the pH of the solution decreases. Therefore, a pH range of 3-5 suggests that the solution is acidic.

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10. -273 āĻĄāĻŋāĻ—ā§āϰ⧀ āϏ⧇āϞāϏāĻŋ⧟āĻžāϏ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ  āύāĻžāχāĻŸā§āϰ⧋āĻœā§‡āύ āĻāϰ āĻŽā§‹āϞāĻžāϰ āĻ†ā§ŸāϤāύ āĻ•āϤ āϞāĻŋāϟāĻžāϰ?

Explanation

not-available-via-ai

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11. HPLC āϤ⧇ āϏāϚāϞ āĻŽāĻžāĻ§ā§āϝāĻŽ āĻšāĻŋāϏ⧇āĻŦ⧇ āĻ•āĻŋ āĻŦā§āϝāĻŦāĻšā§ƒāϤ āĻšā§Ÿ?

Explanation

In HPLC (High Performance Liquid Chromatography), methanol and water are commonly used as the mobile phase. The mobile phase is the solvent or mixture of solvents that carries the sample through the chromatographic system. Methanol and water are often chosen as the mobile phase because they have good solubility, low viscosity, and can effectively elute a wide range of compounds. Additionally, methanol and water can be easily mixed in different ratios to optimize the separation of analytes in the sample. Therefore, methanol and water are commonly used as the mobile phase in HPLC.

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12. āĻāϏāĻŋāĻĄ āĻŽāĻžāĻ§ā§āϝāĻŽā§‡ āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻĒāĻžāϰ āĻŽā§āϝāĻžāĻ™ā§āĻ—āĻžāύ⧇āϟ āĻāϰ āĻ—ā§ƒāĻšā§€āϤ āχāϞ⧇āĻ•āĻŸā§āϰāύ āϏāĻ‚āĻ–ā§āϝāĻž-

Explanation

not-available-via-ai

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13. 12.25 gm Na2CO3 āϕ⧇ āϏāĻŽā§āĻĒā§‚āĻ°ā§āĻŖ āĻĒā§āϰāĻļāĻŽāĻŋāϤ āĻ•āϰāϤ⧇ āĻ•āϤ āĻ—ā§āϰāĻžāĻŽ HCl āϞāĻžāĻ—āĻŦ⧇?

Explanation

To completely neutralize 12.25 gm of Na2CO3, we need to calculate the molar mass of Na2CO3. The molar mass of Na2CO3 is 106 g/mol. Since Na2CO3 is a base and HCl is an acid, they react in a 1:1 ratio. Therefore, we can use the equation:

mass of HCl = (molar mass of Na2CO3) / (molar mass of HCl) * mass of Na2CO3

Plugging in the values, we get:

mass of HCl = (106 g/mol) / (36.5 g/mol) * 12.25 gm

mass of HCl = 8.436 gm

Therefore, 8.436 gm of HCl is required to completely neutralize 12.25 gm of Na2CO3.

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14. 10g CaCO3 āĻĨ⧇āϕ⧇ 2X1020 āϟāĻŋ āĻ…āϪ⧁ āϏāϰāĻŋā§Ÿā§‡ āύāĻŋāϞ⧇ āĻ•āϤāϟ⧁āϕ⧁ CaCOāĻŦāĻžāĻ•āĻŋ āĻĨāĻžāĻ•āĻŦ⧇?

Explanation

If 10g of CaCO3 is decomposed to form 2X10^20 particles, then the molar mass of CaCO3 can be calculated using Avogadro's number. The molar mass of CaCO3 is 100.09 g/mol. So, the number of moles of CaCO3 decomposed is 10g / 100.09 g/mol = 0.0999 mol. Since 2X10^20 particles are formed from 1 mol of CaCO3, the remaining amount of CaCO3 can be calculated by multiplying the number of moles by the molar mass: 0.0999 mol * 100.09 g/mol = 9.966 g. Therefore, the correct answer is 9.966 g.

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15. āĻŽā§ƒāĻĻ⧁ āĻāϏāĻŋāĻĄ āĻ“ āϤ⧀āĻŦā§āϰ āĻ•ā§āώāĻžāϰ⧇āϰ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžā§Ÿ āωāĻĒāϝ⧁āĻ•ā§āϤ āύāĻŋāĻ°ā§āĻĻ⧇āĻļāĻ• āϕ⧋āύāϟāĻŋ?

Explanation

ফেনফথ্যালিন হল একটি মৃদু এসিড ও তীব্র ক্ষারের বিক্রিয়ায় উপযুক্ত নির্দেশক। এটি এসিড-বেস নির্দেশক হিসেবে ব্যবহার করা হয়। এর মাধ্যমে এসিড ও ক্ষারের পরিমাণ নির্ণয় করা যায়। এটি রঙের পরিবর্তন করে এসিড ও ক্ষারের নির্ণয় করার জন্য ব্যবহৃত হয়। সম্পূর্ণ বাক্যটির অর্থ হল, ফেনফথ্যালিন হল মৃদু এসিড ও তীব্র ক্ষারের বিক্রিয়ায় উপযুক্ত নির্দেশক।

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16. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻŦ⧇āύāϜāĻŋāύ āĻĻā§āϰāĻŦāϪ⧇ āĻ•āĻžāĻ°ā§āĻŦāύ⧇āϰ āĻļāϤāĻ•āϰāĻž āϏāĻ āĻŋāĻ• āĻĒāϰāĻŋāĻŽāĻžāĻĒ?

Explanation

not-available-via-ai

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17. 3.5% NaHCO3 āĻāϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻž āĻ•āϤ āĻŽā§‹āϞāĻžāϰ?

Explanation

The molar mass of NaHCO3 (sodium bicarbonate) is 84.01 g/mol. To find the molarity, we need to divide the given mass (3.5%) by the molar mass.

3.5% of NaHCO3 means 3.5 g of NaHCO3 in 100 g of solution.

So, the mass of NaHCO3 is 3.5 g.

Now, we can calculate the moles of NaHCO3 by dividing the mass by the molar mass:

moles = mass / molar mass = 3.5 g / 84.01 g/mol ≈ 0.04167 mol

Therefore, the molarity of 3.5% NaHCO3 is approximately 0.04167 M, which is equivalent to 0.4167 in the given options.

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18. 1.032 g O2 āĻ“ 0.573g CO2 āĻŽāĻŋāĻļā§āϰāϪ⧇āϰ CO2 āĻāϰ āĻŽā§‹āϞ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļ āĻ•āϤ?

Explanation

The molar mass of O2 is 32 g/mol and the molar mass of CO2 is 44 g/mol. To find the mole fraction of CO2 in the mixture, we need to calculate the moles of CO2 and the total moles of the mixture.

The moles of O2 can be calculated by dividing its mass by its molar mass:
moles of O2 = 1.032 g / 32 g/mol = 0.03225 mol

The moles of CO2 can be calculated in the same way:
moles of CO2 = 0.573 g / 44 g/mol = 0.01302 mol

The total moles of the mixture is the sum of the moles of O2 and CO2:
total moles = 0.03225 mol + 0.01302 mol = 0.04527 mol

Finally, the mole fraction of CO2 can be calculated by dividing the moles of CO2 by the total moles:
mole fraction of CO2 = 0.01302 mol / 0.04527 mol = 0.287

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19. āĻŦāĻŋ⧟āĻžāϰ āĻ˛ā§āϝāĻžāĻŽā§āĻŦāĻžāϰāϟ āϏ⧂āĻ¤ā§āϰ āϕ⧋āύ āĻŽā§‹āϞāĻžāϰ āĻĻā§āϰāĻŦāϪ⧇āϰ āĻ•ā§āώ⧇āĻ¤ā§āϰ⧇ āĻ…āϧāĻŋāĻ• āĻĒā§āϰāϝ⧋āĻœā§āϝ?

Explanation

The correct answer is 0.01 because the Beer-Lambert law states that the absorbance of a substance is directly proportional to the concentration of the substance in the solution. Therefore, a smaller value of absorbance (0.01) indicates a lower concentration of the solution, making it more applicable in the context of the Beer-Lambert law.

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20. āϞāϘ⧁ H2SO4 āĻ āĻāĻ• āϟ⧁āĻ•āϰāĻž āϞ⧋āĻšāĻžāϰ āϤāĻžāϰ āĻĻā§āϰāĻŦā§€āĻ­ā§‚āϤ āĻ•āϰāĻžāϰ āĻĒāϰ āĻĒā§āϰāĻžāĻĒā§āϤ āĻĻā§āϰāĻŦāĻŖāϕ⧇ āϏāĻŽā§āĻĒā§‚āĻ°ā§āĻŖ āϜāĻžāϰāĻŋāϤ āĻ•āϰāϤ⧇ 0.03 M KMnO4 āĻĻā§āϰāĻŦāϪ⧇āϰ 27.5 mL āϞāĻžāϗ⧇āĨ¤ āϞ⧋āĻšāĻžāϰ āϤāĻžāϰ⧇āϰ āĻ­āϰ āĻ•āϤ?

Explanation

The given question states that 27.5 mL of 0.03 M KMnO4 solution is required to completely oxidize a piece of iron in dilute H2SO4. From this information, we can calculate the number of moles of KMnO4 used. Then, using the balanced chemical equation, we can determine the number of moles of iron that reacted. Finally, using the molar mass of iron, we can calculate its mass. The correct answer is 0.23 g, which is the mass of iron required to react with the given amount of KMnO4.

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21. 60g āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻ•ā§āϞ⧋āϰ⧇āϟāϕ⧇ āωāϤāĻĒā§āϤ āĻ•āϰ⧇ āĻĒā§āϰāĻžāĻĒā§āϤ āĻ…āĻ•ā§āϏāĻŋāĻœā§‡āύ āĻ—ā§āϝāĻžāϏ⧇āϰ āĻ†ā§ŸāϤāύ NTP āϤ⧇ āĻ•āϤ āϞāĻŋāϟāĻžāϰ?

Explanation

not-available-via-ai

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22. 1.881 gm āϭ⧇āϜāĻžāϞāϝ⧁āĻ•ā§āϤ Na2CO3 āϕ⧇ āĻĒāĻžāύāĻŋāϤ⧇ āĻĻā§āϰāĻŦā§€āĻ­ā§‚āϤ āĻ•āϰ⧇ āĻ†ā§ŸāϤāύ 250 cc āĻ•āϰāĻž āĻšāϞāĨ¤ āĻ āĻĻā§āϰāĻŦāϪ⧇āϰ 25 cc, M/10 āĻŽāĻžāĻ¤ā§āϰāĻžāϰ 24.05 cc HCl āĻĻā§āϰāĻŦāϪ⧇ āĻĒā§‚āĻ°ā§āĻŖ āĻĒā§āϰāĻļāĻŽāĻŋāϤ āĻ•āϰāϞ⧇ āĻĻā§āϰāĻŦāϪ⧇ āϭ⧇āϜāĻžāϞ⧇āϰ āĻļāϤāĻ•āϰāĻž āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻ•āϤ?

Explanation

When 25 cc of the solution is titrated with 24.05 cc of M/10 HCl, it indicates that 1 cc of M/10 HCl is required to neutralize 0.881 gm of Na2CO3. Therefore, to neutralize 1 gm of Na2CO3, (24.05/25) * 0.881 = 0.8464 gm of HCl is required. The molar mass of HCl is 36.5 gm/mol, so the molarity of HCl is (0.8464/36.5) mol/L. As the volume of the solution is 250 cc, the amount of Na2CO3 in the solution is (0.8464/36.5) * 250 = 5.853 grams. Therefore, the percentage of Na2CO3 in the solution is (5.853/18.02) * 100 = 32.23%.

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23. 250 mL āĻĻā§āϰāĻŦāϪ⧇  12.75 gm K2Cr2O7 āĻĨāĻžāĻ•āϞ⧇ āĻĻā§āϰāĻŦāϪ⧇āϰ āĻŽā§‹āϞāĻžāϰāĻŋāϟāĻŋ āĻ•āϤ āĻŽā§‹āϞāĻžāϰ?

Explanation

The molar mass of K2Cr2O7 is 294.18 g/mol. To find the molarity, we need to divide the given mass (12.75 g) by the molar mass.

Molarity = mass/molar mass
Molarity = 12.75 g/294.18 g/mol
Molarity ≈ 0.043 mol/L

However, the question asks for the molarity in moles, not moles per liter. Since the volume of the solution is given as 250 mL, we need to convert it to liters by dividing by 1000.

Molarity = 0.043 mol/0.25 L
Molarity ≈ 0.17 mol

Therefore, the molarity of the solution is approximately 0.17 M.

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24. 0.1 N H2SO4 āĻĻā§āϰāĻŦāϪ⧇āϰ āĻŽā§‹āϞāĻžāϰāĻŋāϟāĻŋ āĻ•āϤ āĻšāĻŦ⧇?

Explanation

The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the question asks for the molarity of a 0.1 N H2SO4 solution. N stands for normality, which is a measure of the number of equivalents of a solute per liter of solution. Since 0.1 N is equivalent to 0.1 moles per liter, the molarity of the solution will be M/10.

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25. āĻĒāĻžāύāĻŋāϰ āϘāύāĻ¤ā§āĻŦ 0.9987 gm/cc āĻ“ āχāωāϰāĻŋ⧟āĻžāϰ āφāĻŖāĻŦāĻŋāĻ• āĻ­āϰ 60 āĻšāϞ⧇ 50 g āχāωāϰāĻŋ⧟āĻž 850 g āĻĒāĻžāύāĻŋāϤ⧇ āĻĻā§āϰāĻŦā§€āĻ­ā§‚āϤ āĻ•āϰāϞ⧇ āĻĻā§āϰāĻŦāϪ⧇āϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻž-

Explanation

When 50 g of urea is dissolved in 850 g of water, the volume of the solution is the sum of the volumes of urea and water. Since density is mass divided by volume, we can calculate the volume of the solution by dividing the mass of the solution by its density. The mass of the solution is 50 g + 850 g = 900 g. Therefore, the volume of the solution is 900 g / 0.9987 gm/cc = 900 cc. Now, we can calculate the molarity of the solution by dividing the moles of urea by the volume of the solution in liters. The moles of urea is calculated by dividing the mass of urea by its molar mass. The molar mass of urea (CH4N2O) is 60 g/mol. Therefore, the moles of urea is 50 g / 60 g/mol = 0.8333 mol. Finally, we can calculate the molarity by dividing the moles of urea by the volume of the solution in liters: 0.8333 mol / 0.9 L = 0.925 M. However, none of the given options match this value, so the correct answer is not available.

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26. 9.15 g āĻŦāĻŋāĻļ⧁āĻĻā§āϧ ZnO āĻšāϤ⧇ āĻ•āĻžāĻ°ā§āĻŦāύ āĻŦāĻŋāϜāĻžāϰāĻŖ āĻĒāĻĻā§āϧāϤāĻŋāϤ⧇ āĻ•āϤ āĻ—ā§āϰāĻžāĻŽ Zn āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ?

Explanation

The molar mass of ZnO is 81.38 g/mol. In the carbon reduction method, carbon reacts with ZnO to produce Zn and CO. The balanced equation for the reaction is: ZnO + C -> Zn + CO. According to the equation, 1 mole of ZnO reacts with 1 mole of C to produce 1 mole of Zn. Therefore, the number of moles of Zn can be calculated by dividing the given mass of ZnO (9.15 g) by its molar mass (81.38 g/mol). The result is approximately 0.1125 moles. Since the molar ratio between ZnO and Zn is 1:1, the number of moles of Zn is also 0.1125. Finally, the mass of Zn can be calculated by multiplying the number of moles of Zn by its molar mass (65.38 g/mol), which gives a result of approximately 7.35 g.

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27. āĻŦāĻžāĻˇā§āĻĒ āĻĨ⧇āϕ⧇ 2 gm H2 āĻĒā§āϰāĻ¸ā§āϤ⧁āϤ āĻ•āϰāϤ⧇ āĻ•āĻŋ āĻĒāϰāĻŋāĻŽāĻžāĻŖ Fe āĻāϰ āĻĒā§āĻ°ā§Ÿā§‹āϜāύ āĻšā§Ÿ?

Explanation

not-available-via-ai

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28. 1L āϏāĻŽā§āĻĻā§āϰ⧇āϰ āĻĒāĻžāύāĻŋ āύāĻŋā§Ÿā§‡ āĻŦāĻžāĻˇā§āĻĒā§€āϭ⧁āϤ āĻ•āϰ⧇ 38.5 g āĻļ⧁āĻˇā§āĻ• āϞāĻŦāĻŖ āĻĒāĻžāĻ“ā§ŸāĻž āϗ⧇āϞāĨ¤ āϏāĻŽā§āĻĻā§āϰ⧇āϰ āĻĒāĻžāύāĻŋāϰ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āϗ⧁āϰ⧁āĻ¤ā§āĻŦ 1.03 āĻšāϞ⧇ āĻāϤ⧇ āĻ•āĻ āĻŋāύ āĻŦāĻ¸ā§āϤ⧁āϰ āĻļāϤāĻ•āϰāĻž āĻšāĻžāϰ āĻ•āϤ?

Explanation

The relative density of a substance is the ratio of its density to the density of water. In this question, the relative density of the sea water is given as 1.03. The amount of dry salt obtained from 1L of sea water is 38.5g. To find the percentage of salt in sea water, we divide the mass of salt by the mass of water and multiply by 100. Therefore, (38.5g/1000g) x 100 = 3.85%. The relative density of the salt is then calculated by dividing the percentage of salt by 100. So, 3.85/100 = 0.0385. Finally, to find the specific gravity, we divide the relative density of the salt by the relative density of water. Hence, 0.0385/1.03 = 0.0374 or 3.74%.

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29. āĻ•āĻĒāĻžāϰ āϏāĻžāϞāĻĢ⧇āϟ āĻ“ āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻ†ā§Ÿā§‹āĻĄāĻžāχāĻĄā§‡āϰ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžā§Ÿ 1 mol āĻ†ā§Ÿā§‹āĻĄāĻŋāύ āϤ⧈āϰāĻŋāϤ⧇ āĻ•āϤ āĻ—ā§āϰāĻžāĻŽ āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻ†ā§Ÿā§‹āĻĄāĻžāχāĻĄ āĻĒā§āĻ°ā§Ÿā§‹āϜāύ?

Explanation

In the given reaction, 1 mole of iodine is produced by the reaction between copper sulfate and potassium iodide. To find the amount of potassium iodide needed, we need to calculate the molar mass of potassium iodide (KI). The molar mass of KI is 39.1 g/mol (for K) + 126.9 g/mol (for I) = 166 g/mol. Therefore, to produce 1 mole of iodine, we need 166 g of potassium iodide. However, the question asks for the amount of potassium iodide needed to produce 1 mole of iodine, which is equivalent to 2 moles of potassium iodide. Therefore, we multiply the molar mass of KI by 2, which gives us 332 g. Hence, the correct answer is 332 g.

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30. āĻŦāĻžāĻ—āĻžāύ⧇ āĻŦā§āϝāĻŦāĻšā§ƒāϤ āϏāĻžāϰ⧇ 30% āĻĢāϏāĻĢāϰāĻžāϏ P2O5 āĻšāĻŋāϏ⧇āĻŦ⧇ āĻĨāĻžāĻ•āϞ⧇ āϐ āϏāĻžāϰ⧇ āĻĢāϏāĻĢāϰāĻžāϏ⧇āϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻ•āϤ?

Explanation

If 30% of the fertilizer used in the garden is in the form of P2O5, it means that 30% of the fertilizer is composed of phosphorus. Since P2O5 contains 44.3% phosphorus, we can calculate the amount of phosphorus in the fertilizer by multiplying 30% by 44.3%. This gives us 13.1%, which means that 13.1% of the fertilizer is composed of phosphorus. Therefore, the correct answer is 13.1%.

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āύāĻŋāĻšā§‡āϰ āϕ⧋āύ āĻ†ā§Ÿāύ āϜāĻžāϰāĻ• āĻ“...
āĻ—ā§āϝāĻžāϏ āĻ•ā§āϰ⧋āĻŽāĻžāĻŸā§‹āĻ—ā§āϰāĻžāĻĢāĻŋāϤ⧇...
Ca(OCl)Cl āĻ…āϪ⧁āϤ⧇ Cl āĻāϰ āϜāĻžāϰāĻŖ āĻ…āĻŦāĻ¸ā§āĻĨāĻž-
āĻĻā§āϰāĻŦāϪ⧇āϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻžāϰ āϕ⧋āύ āĻāĻ•āĻ•...
āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻĒā§āϰāĻžāχāĻŽāĻžāϰ⧀...
5g FeSO4 āϕ⧇ āĻ…āĻŽā§āĻ˛ā§€ā§Ÿ āĻŽāĻžāĻ§ā§āϝāĻŽā§‡...
5% āϭ⧇āϜāĻžāϞāϝ⧁āĻ•ā§āϤ 100 g āϚ⧁āύāĻžāĻĒāĻžāĻĨāϰ...
0.1 M āϏāĻŽāĻ†ā§ŸāϤāύ⧇āϰ NaOH āĻ“ H2SO4 āĻāϰ...
āĻŽāĻŋāĻĨāĻžāχāϞ āĻ…āϰ⧇āĻžā§āĻœā§‡āϰ āĻŦāĻ°ā§āĻŖ...
-273 āĻĄāĻŋāĻ—ā§āϰ⧀ āϏ⧇āϞāϏāĻŋ⧟āĻžāϏ...
HPLC āϤ⧇ āϏāϚāϞ āĻŽāĻžāĻ§ā§āϝāĻŽ āĻšāĻŋāϏ⧇āĻŦ⧇ āĻ•āĻŋ...
āĻāϏāĻŋāĻĄ āĻŽāĻžāĻ§ā§āϝāĻŽā§‡ āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻĒāĻžāϰ...
12.25 gm Na2CO3 āϕ⧇ āϏāĻŽā§āĻĒā§‚āĻ°ā§āĻŖ...
10g CaCO3 āĻĨ⧇āϕ⧇ 2X1020 āϟāĻŋ āĻ…āϪ⧁ āϏāϰāĻŋā§Ÿā§‡...
āĻŽā§ƒāĻĻ⧁ āĻāϏāĻŋāĻĄ āĻ“ āϤ⧀āĻŦā§āϰ āĻ•ā§āώāĻžāϰ⧇āϰ...
āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻŦ⧇āύāϜāĻŋāύ āĻĻā§āϰāĻŦāϪ⧇...
3.5% NaHCO3 āĻāϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻž āĻ•āϤ...
1.032 g O2 āĻ“ 0.573g...
āĻŦāĻŋ⧟āĻžāϰ āĻ˛ā§āϝāĻžāĻŽā§āĻŦāĻžāϰāϟ āϏ⧂āĻ¤ā§āϰ...
āϞāϘ⧁ H2SO4 āĻ āĻāĻ• āϟ⧁āĻ•āϰāĻž āϞ⧋āĻšāĻžāϰ...
60g āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ āĻ•ā§āϞ⧋āϰ⧇āϟāϕ⧇...
1.881 gm āϭ⧇āϜāĻžāϞāϝ⧁āĻ•ā§āϤ Na2CO3 āϕ⧇...
250 mL āĻĻā§āϰāĻŦāϪ⧇  12.75 gm K2Cr2O7 āĻĨāĻžāĻ•āϞ⧇...
0.1 N H2SO4 āĻĻā§āϰāĻŦāϪ⧇āϰ āĻŽā§‹āϞāĻžāϰāĻŋāϟāĻŋ āĻ•āϤ...
āĻĒāĻžāύāĻŋāϰ āϘāύāĻ¤ā§āĻŦ 0.9987 gm/cc āĻ“ āχāωāϰāĻŋ⧟āĻžāϰ...
9.15 g āĻŦāĻŋāĻļ⧁āĻĻā§āϧ ZnO āĻšāϤ⧇ āĻ•āĻžāĻ°ā§āĻŦāύ...
āĻŦāĻžāĻˇā§āĻĒ āĻĨ⧇āϕ⧇ 2 gm H2 āĻĒā§āϰāĻ¸ā§āϤ⧁āϤ...
1L āϏāĻŽā§āĻĻā§āϰ⧇āϰ āĻĒāĻžāύāĻŋ āύāĻŋā§Ÿā§‡...
āĻ•āĻĒāĻžāϰ āϏāĻžāϞāĻĢ⧇āϟ āĻ“ āĻĒāϟāĻžāĻļāĻŋ⧟āĻžāĻŽ...
āĻŦāĻžāĻ—āĻžāύ⧇ āĻŦā§āϝāĻŦāĻšā§ƒāϤ āϏāĻžāϰ⧇ 30%...
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