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

A solution of hydrochloric acid in water is 38.00%hydrochloric acid by mass. Its density is 1.1886gcm-3. Compute its molarity, mole fraction, and molality at this temperature.

Short Answer

Expert verified

Step by step solution

01

Define molarity, molality and mole fraction of a solution

02

Calculate the number of moles of hydrochloric acid and water

03

Determine the mole fraction of the solution

04

Calculate the volume of hydrochloric acid present in the solution

05

Determine the molarity of the solution

06

Determine the molality of the solution

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!

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

Complete and balance the following equations for reactions taking place in an acidic solution.

(a)VO2+(aq)+ SO2(g)VO2 +(aq) + SO42 -(aq)

(b)Br2(l) + SO2(g)Br-(aq) + SO42 -(aq)

(c)Cr2O72 -(aq) + Np4 +(aq)Cr3 +(aq) + NpO22 +(aq)

(d)HCOOH(aq) + MNO4-(aq)CO2(g) + Mn2 +(aq)

(e)Hg2HPO4(s) + Au(s) + Cl-(aq)Hg(l) +H2PO4-(aq) + AuCl4-(aq)

The amount of ozone in a mixture of gases can be determined by passing the mixture through an acidic aqueous solution of potassium iodide, where the ozone reacts according to

O3(g)+3I-(aq)+H2O()O2(g)+I3-(aq)+2OH-(aq)

to form the triiodide ionI3-. The amount of triiodide produced is then determined by titrating with thiosulfate solution:

I3-(aq)+2S2O32-(aq)3I-(aq)+S4O62-(aq)

The solution requires 26.2 mL of a 0.1359-M solution of thiosulfate ion to titrate to the endpoint. Calculate the mole fraction of ozone in the original gas sample. A small amount of starch solution is added as an indicator because it forms a deep-blue complex with the triiodide solution. Disappearance of the blue color thus signals the completion of the titration. Suppose 53.2 L of a gas mixture at a temperature of18°Cand a total pressure of 0.993 atm is passed through a solution of potassium iodide until the ozone in the mixture has reacted completely. The solution requires 26.2 mL of a 0.1359-M solution of thiosulfate ion to titrate to the endpoint. Calculate the mole fraction of ozone in the original gas sample.

Vanadic ion,V3+ , forms green salts and is a good reacting agent, being itself changed in neutral solutions to the nearly colorless ionVOH4+ . Suppose that 15.0 mL of a0.200M solution of vanadic sulfate,V2SO43 , was needed to reduce completely a 0.540gsample of an unknown substanceX . If each molecule of Xaccepted just one electron, what is the molecular weight ofX ? Suppose that each molecule of Xaccepted three electrons; what would be the molecular weight of Xthen?

You take a bottle of soft drink out of your refrigerator. The contents are liquid and stay liquid, even when you shake to water. Presently, you remove the cap and the liquid freezes solid. Offer a possible explanation for this observation.

At 250C , some benzene (C6H6)is added to a sample of gascous methane CH4at 1.00 atmpressure in a closed vessel, and the vessel is shaken until as much methane as possible dissolves. Then 1.00 kg of the solution is removed and boiled to expel the methane, yielding the volume of methane that results is 0.510 L atrole="math" localid="1663672111295" 00C and 1.00 atm . Determine the Henry's law constant for methane in benzene.

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