Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The gas that turns lime water milky is (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{CO}_{2}\) (c) both of these (d) none of these

Short Answer

Expert verified
(b) \( \mathrm{CO}_2 \)

Step by step solution

01

Understand the Reaction

When a specific gas is bubbled through lime water, a chemical reaction takes place that can cause the water to turn milky. This reaction is a key indicator of the presence of a certain gas.
02

Identify Lime Water Composition

Lime water is a saturated solution of calcium hydroxide, \ \( \mathrm{Ca(OH)}_2 \ \), in water. Its interaction with certain gases can lead to the formation of calcium carbonate, which is insoluble in water and thus appears milky.
03

Examine Each Option

Consider each gas: \ (a) \ \( \mathrm{SO}_2 \ \) - Sulfur dioxide does not typically react with lime water to form a precipitate.(b) \ \( \mathrm{CO}_2 \ \) - Carbon dioxide reacts with \ \( \mathrm{Ca(OH)}_2 \ \) to form \ \( \mathrm{CaCO}_3 \ \), which is insoluble and makes the water appear milky.(c) Both of these - Only \ \( \mathrm{CO}_2 \ \) has the described effect on lime water.(d) None of these - Eliminated based on the reaction of \ \( \mathrm{CO}_2 \ \).
04

Conclude the Correct Answer

Based on the steps above, only carbon dioxide \ \( \mathrm{CO}_2 \ \) turns lime water milky. Therefore, the correct answer is (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lime Water Test
The Lime Water Test is a simple yet effective method used to identify the presence of carbon dioxide in a sample of gas. It is widely employed in chemistry labs due to its straightforward procedure and clear visible results.
When you have a solution called lime water, which is calcium hydroxide dissolved in water, its interaction with gases can be very telling.
In this test, if carbon dioxide gas is bubbled through lime water, a chemical reaction occurs that produces calcium carbonate, a milky-looking compound. This reaction is a specific indicator of carbon dioxide:
  • The lime water test is used as a quick check for CO₂.
  • The resulting milky appearance is due to the formation of a solid compound that doesn’t dissolve in water.
  • It’s a key method for detecting carbon dioxide in various settings.
For students learning chemistry, understanding this test is crucial because it demonstrates one of the fundamental chemical reactions undergone by carbon dioxide.
Calcium Hydroxide
Calcium hydroxide, known chemically as \(\text{{Ca(OH)}}_2\), is a chemical compound that plays a significant role in various chemical reactions including the lime water test. It is often referred to as "slaked lime" and is a widely used material in both industrial and laboratory settings.
This compound is slightly soluble in water, creating a solution called lime water when dissolved. The nature of calcium hydroxide makes it particularly useful in the reaction with carbon dioxide, resulting in calcium carbonate.
Some important points about calcium hydroxide include:
  • It is derived from calcium oxide through the addition of water.
  • The solution of calcium hydroxide (lime water) is clear until it reacts with carbon dioxide.
  • In its solid state, it appears as a white powder.
Knowing about calcium hydroxide helps us understand why lime water reacts as it does when exposed to certain gases like carbon dioxide.
Carbon Dioxide Reaction
When carbon dioxide (CO₂) comes into contact with lime water, it reacts with calcium hydroxide to form calcium carbonate and water. This specific chemical reaction is crucial because it provides a visual confirmation of the presence of carbon dioxide.
The equation for this reaction is:\[\text{{Ca(OH)}}_2(aq) + \text{{CO}}_2(g) \rightarrow \text{{CaCO}}_3(s) + \text{{H}}_2\text{{O}}(l)\]
Here's what's happening in the reaction:
  • Calcium hydroxide in the lime water reacts with carbon dioxide from the gas to form a precipitate.
  • The calcium carbonate formed is not soluble in water, thus giving the solution a cloudy, milky appearance.
  • This reaction confirms the presence of carbon dioxide conclusively.
Understanding this reaction is vital for chemistry students, as it provides insight into the behavior of carbon-based compounds and their interactions.
Sulfur Dioxide
Sulfur dioxide (SO₂) is a gas that, unlike carbon dioxide, does not react with lime water to form a precipitate or cause the solution to turn milky. This difference is important to note, as confusing these gases could lead to incorrect conclusions in experiments.
While sulfur dioxide does participate in various reactions, its interaction with lime water is distinct from that of carbon dioxide:
  • Unlike CO₂, SO₂ does not produce calcium carbonate when bubbled through lime water.
  • The absence of a milky appearance indicates the presence of SO₂ rather than CO₂.
  • Sulfur dioxide's typical reactions often involve oxidation rather than precipitation.
Knowing how sulfur dioxide behaves compared to carbon dioxide can save students from making errors in interpreting simple yet significant chemical tests.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which is incorrect here? (1) Heat of neutralization for strong acid and strong base is always less than \(13.7 \mathrm{~kJ}\). (2) Aniline yellow can also be used to prepare other dyes. (3) Phenolphthalein can be used as an indicator in the titration of strong acid and weak base (4) Methyl orange works in the range of \(3.1-4.4\). (a) 2 and 4 (b) 2 and 3 (c) 1 and 3 (d) 3 and 4

Identify the products \(\mathrm{A}\) and \(\mathrm{B}\) here. (1) \(2 \mathrm{Ag}^{+}\)(excess) \(+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) \(\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{A}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(2 \mathrm{NO}_{2}^{-}+2 \mathrm{I}^{-} \stackrel{\text { acid medium }}{\longrightarrow} \mathrm{B}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=\mathrm{N}_{2}\) (b) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=2 \mathrm{NO}_{2}\) (d) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=2 \mathrm{NO}\)

A metal \(X\) on heating in nitrogen gas gives \(Y . Y\) on treatment with \(\mathrm{H}_{2} \mathrm{O}\) gives a colourless gas which when passed through \(\dot{\mathrm{CuSO}_{4}}\) solution gives a blue colour. Y is (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) (b) \(\mathrm{MgO}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Which is correct statement here? (1) Glacial acetic acid is used as a solvent in the preparation of p-nitroacetanilide. (2) Potash alum is used as a mordant in dyeing. (3) Potassium alum is used to remove positively charged impurities from water. (4) Iodoform has antiseptic properties also. (a) 2 and 4 (b) 1,3 and 4 (c) 1 and 3 (d) 1,2 and 4

A certain pale-green substance, \(X\) becomes dark brown on adding \(\mathrm{NaNO}_{2}\) in presence of dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\). its aqueous solution gives precipitates with (i) \(\mathrm{BaCl}_{2}\) and (ii) \(\mathrm{NaOH}\) in separate tests. The latter \(\mathrm{NaOH}\) in separate tests. The latter precipitate, Y, gradually changes colour from green to brown, on exposure to air. Identify X here? (a) \(\mathrm{NiSO}_{4}\) (b) \(\mathrm{FeSO}_{4}\) (c) \(\mathrm{ZnSO}_{4}\) (d) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free