Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Q5TY (page 443)

Dilution by mass is more accurate than dilution by volume. A stocksolution contains 1.044 g Fe/kg solution in 0.48 M HCl. How many μgFe/gsolution are in a solution made by mixing 2.145 g stock solution with 243.27 g 0.1 M HCl?

Short Answer

Expert verified

The amount ofμgFe/gis9.125μgFe/g

Step by step solution

01

Calculate the Mass:

AmountFe(μg/g)=MassofFeMassofsolution.......(1)

Equation for mass of solution,

Mass of solution = Mass of stock solution + Mass of HCL

Given,

Mass of stock solution =2.145 g

Mass of HCL=243.27 g

Mass of solution =2.145gn+243.27g

Massofsolution=245.415g

Mass of Fe in stock solution = (Mass of Fe per kg solution)(Mass of stock solution)

Mass of Fe in stock solution = 1.044g1kg1kg103g

Mass of Fe in stock solution =0.00223938g

Substitute in equation (1),

02

STEP:2 :  Calculate  the amount of iron (Fe)

AmountofFe(μg/g)=MassofFeMassofsolution

role="math" localid="1663669292492" AmountofFe(μg/g)=0.00223938245.415g1061gAmountofFe(μg/g)≈9.125μg/g

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.

Start your free trial

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

Gold nanoparticles (Figure 17-31) can be titrated with the oxidizing agent TCNQ in the presence of excess ofBr2-to oxidize Au(0)"toAuBr2-in deaerated toluene. Gold atoms in the interior of the particle are Au(0) . Gold atoms bound to C12H25S- (dodecanethiol) ligands on the surface of the particle are Au(I) and are not titrated.

The table gives the absorbance at $856 0.700 mL of 1.00×10-4MTCNQ+0.05M(C8H17)4N+Br-in toluene is titrated with gold nanoparticles (1.43 g/Lin toluene) from a microsyringe with a Teflon-coated needle. Absorbance in the table has already been corrected for dilution.

(a) Make a graph of absorbance versus volume of titrant and estimate the equivalence point. Calculate the Au(0) in 1.00 g of nanoparticles.

(b) From other analyses of similarly prepared nanoparticles, it is estimated that 25 % of the mass of the particle is dodecanethiol ligand. Calculate mmol of localid="1667559229564" C12H25Sin 1.00 g of nanoparticles.

(c) The Au(I) content of 1.00 g of nanoparticles should be 1.00 mass of Au(0) - mass of C12H25S. Calculate the micromoles of Au(I) in 1.00 g of nanoparticles and the mole ratio Au(I) :. In principle, this ratio should be 1: 1. The difference is most likely because C12H25Swas not measured for this specific nanoparticle preparation.

A compound with molecular mass 292.16g/molwas dissolved in a 5 - mLvolumetric flask. A1.00 - mLaliquot was withdrawn, placed in a 10 - mLvolumetric flask, and diluted to the mark. The absorbance atwas in 0.427 i acuvet. The molar absorptivity at 340nm isε340=6130M-1cm-1

(a) Calculate the concentration of compound in the cuvet.

(b) What was the concentration of compound in the 5 - mL flask?

(c) How many milligrams of compound were used to make the 5 - mLsolution?

You have been sent to India to investigate the occurrence ofgoiter disease attributed to iodine deficiency. As part of your investigation, you must make field measurements of traces of iodide I2in groundwater. The procedure is to I2 into an intensely colored complex with the dye brilliant green inthe organic solvent toluene.

(a)A 3.15×10-6Msolution of the colored complex exhibited an absorbance of 0.267 at 635nmin a 1.00 - cm cuvet. A blank solution made from distilled water in place of groundwater had an absorbance of 0.019 . Find the molar absorptivity of the coloredcomplex.

(b) The absorbance of an unknown solution prepared from groundwater was0.175. Find the concentration of the unknown.

(a) A 3.96×10-4Msolution of compound A exhibited an absorbance of 0.624 at 238nmin a 1.000 cmcuvet; a blank solution containing only solvent had an absorbance of 0.029at the same wavelength. Find the molar absorptivity of compound A.

(b) The absorbance of an unknown solution of compound A in the same solvent and cuvet was 0.375 at 238 nm. Find the concentration of A in the unknown.

(c) A concentrated solution of compound A in the same solvent was diluted from an initial volume ofto a final volume of 25.00mLand then had an absorbance of 0.733. What is the concentration of A in the concentrated solution?

Ammonia can be determined spectrophotometrically by reaction with phenol in the presence of hypochlorite(OCl-)

A 4.37 - mgsample of protein was chemically digested to convert its nitrogen into ammonia and then diluted to 100.0mL. Thenof the solution were placed in a 50 - mLvolumetric flask and treated with 5mLof phenol solution plus 2mLof sodium hypochlorite solution. The sample was diluted to 50.0mLand the absorbance at 625nmwas measured in a 1.00 - cmcuvet after 30min. For reference, a standard solution was prepared from 0.0100g of NH4Cldissolved in 1.00Lof water. Then 10.0mLof this standard were placed in a50 - mL volumetric flask and analyzed in the same manner as the unknown. A reagent blank was prepared by using distilled water in place of unknown.

(a)Calculate the molar absorptivity of the blue product.

(b)Calculate the weight percent of nitrogen in the protein.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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