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One litre solution containing \(490 \mathrm{~g}\) of sulphuric acid is diluted to 10 litres with water. What is the normality of the resulting solution? a. \(1.0 \mathrm{~N}\) b. \(10.0 \mathrm{~N}\) c. \(0.5 \mathrm{~N}\) d. \(5.0 \mathrm{~N}\)

Short Answer

Expert verified
The normality of the resulting solution is \(1.0 \mathrm{~N}\).

Step by step solution

01

Determine the Equivalent Weight of Sulfuric Acid

First, we need to find the equivalent weight of sulfuric acid (H₂SO₄). The molecular weight of H₂SO₄ is calculated as \(2(1) + 32 + 4(16) = 98\) grams per mole. Since sulfuric acid is a diprotic acid, it can donate 2 hydrogen ions per molecule, giving it an equivalent weight of \(\frac{98}{2} = 49\) grams per equivalent.
02

Calculate the Number of Equivalents in Original Solution

The number of equivalents in the original solution is calculated by dividing the weight of sulfuric acid by its equivalent weight: \(\frac{490 \text{ g}}{49 \text{ g/equiv}} = 10 \text{ equivalents}\).
03

Determine the Volume of the Final Solution

The final volume after dilution is given as 10 litres.
04

Calculate Normality of the Solution

The normality of a solution is given by the number of equivalents divided by the volume of the solution in litres. The normality of the resulting solution is: \(\frac{10 \text{ equivalents}}{10 \text{ litres}} = 1.0 \text{ N}\).
05

Choose the Correct Answer

From the calculated normality, we see that the correct choice is (a) \(1.0 \mathrm{~N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulfuric Acid
Sulfuric acid is a strong mineral acid with the chemical formula \(\text{H}_2\text{SO}_4\). It is one of the most utilized acids in various industrial processes. Known for being a diprotic acid, sulfuric acid can donate two protons (hydrogen ions) per molecule in a reaction with a base. This characteristic makes it essential in processes that require strong acidic behavior.
  • It’s typically found in car batteries, fertilizers, and cleaning agents.
  • In aqueous solution, sulfuric acid fully dissociates, which means it splits into its constituent ions readily, making it highly reactive.
  • Its ability to donate two protons classifies it as a strong acid.
Understanding sulfuric acid's behavior is crucial in many chemical reactions, especially those involving neutralization and titration.
Equivalent Weight
The equivalent weight in chemistry is a concept that simplifies the calculation of reacting quantities. For acids, the equivalent weight is determined by dividing the molecular weight by the number of protons the acid can donate. For sulfuric acid, the molecular weight is approximately \(98 \text{ g/mol}\). Since it is diprotic, it can release two protons. Thus, the equivalent weight of sulfuric acid is calculated by dividing its molecular weight by 2, resulting in an equivalent weight of \(49 \text{ g/equiv}\).
  • This metric is crucial for determining how much of an acid is needed to react with a given amount of base in neutralization reactions.
  • It is also essential for calculations involving normality, a measure of concentration that considers the equivalent factor of the solute.
Dilution Process
The dilution process involves reducing the concentration of a solute in a solution, typically by adding more solvent. In the case of sulfuric acid, dilution is often necessary to handle or use the solution safely due to its strong corrosive nature. The exercise you're working on involves diluting a solution of sulfuric acid by adding water until the solution volume reaches ten liters.
  • When diluting, the amount of solute (in equivalents) remains constant, while the volume of the solution increases.
  • This results in a decreased concentration of the original solute, expressed in terms of normality or molarity.
  • Calculating the final concentration, like normality, requires knowing the initial concentration and final volume.
This concept is essential in laboratory and industrial settings where exact concentrations are critical for reactions.
Acid-Base Chemistry
Acid-base chemistry is a fundamental concept in chemistry, focusing on the reactions between acids and bases. These reactions involve the transfer of protons (H⁺ ions) between the reactants. Sulfuric acid is a classic example of a strong acid used in these types of reactions. In the case of sulfuric acid, its two dissociable protons allow for two stages of proton transfer, enhancing its ability in acid-base chemistry.
  • These reactions are usually accompanied by a change in pH, which is a measure of the hydrogen ion concentration in a solution.
  • Strong acids and bases are often fully ionized in solution, creating clear stoichiometry for reactions.
  • The fundamental principles of acid-base chemistry are applied in titration, buffer solutions, and even metabolic processes in the human body.
Understanding acid-base interactions is crucial for predicting and explaining the behavior of these substances in various environments.

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Most popular questions from this chapter

Cobalt (III) ion forms many compounds with ammonia. To find the formula of one of these compounds, you t itrate the \(\mathrm{NH}_{3}\) in the compound with standardized acid. $$ \begin{aligned} &\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}} \mathrm{Cl}_{3}(\mathrm{aq})+{ }_{\mathrm{x}} \mathrm{HCl}(\mathrm{aq}) \rightarrow \\ &{ }_{\mathrm{x}} \mathrm{NH}_{4}{ }^{+}(\mathrm{aq})+\mathrm{Co}^{3+}(\mathrm{aq})+(\mathrm{x}+3) \mathrm{Cl}^{-}(\mathrm{aq}) \end{aligned} $$ Assume that \(23.63 \mathrm{~m} 1\) of \(1.500 \mathrm{M} \mathrm{HCl}\) is used to titrate \(1.580 \mathrm{~g}\) of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}} \mathrm{Cl}_{3} .\) What is the value of \(X ?\) a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) b. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{cl}_{3}\) c. \(\left[\mathrm{Co}(\mathrm{NH})_{3} \mathrm{Cl}_{3}\right.\) d. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\)

A mixture of \(0.002\) mole of \(\mathrm{KBrO}_{3}\) and \(0.01\) mole of \(\mathrm{KBr}\) was treated with excess of KI and acidified. The volume of \(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution required to consume the liberated iodine will be a. \(150 \mathrm{ml}\) b. \(90 \mathrm{ml}\) c. \(120 \mathrm{ml}\) d. \(100 \mathrm{ml}\)

The crystalline salt \(\mathrm{Na}_{2} \mathrm{SO}_{4} \times \mathrm{xH}_{2} \mathrm{O}\) on heating loses \(55.9 \%\) of its weight. The formula of the crystalline salt is a. \(\mathrm{Na}_{2} \mathrm{SO}_{4}, 10 \mathrm{H}_{2} \mathrm{O}\) b. \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{Na}_{2} \mathrm{SO}_{4} .7 \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{Na}_{2} \mathrm{SO}_{4} .5 \mathrm{H}_{2} \mathrm{O}\) e. \(\mathrm{Na}_{2} \mathrm{SO}_{s} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

A metal oxide has the formula \(\mathrm{Z}_{2} \mathrm{O}_{3}\). It can be reduced by hydrogen to give free metal and water \(0.1596 \mathrm{~g}\) of the metal oxide requires \(6 \mathrm{mg}\) of hydrogen for complete reduction. The atomic weight of the metal is a. \(27.9\) b. \(159.6\) c. \(79.8\) d. \(55.8\)

\(0.532\) gram of chloroplatinate of an organic base (mol. wt. 244) gave \(0.195\) gram of platinum on ignition. The number of nitrogen atoms per molecule of base is a. 4 b. 3 c. 2 d. 1

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