/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q3TY 0.10 mM KMnO4has an absorbance... [FREE SOLUTION] | ÷ÈÓ°Ö±²¥

÷ÈÓ°Ö±²¥

0.10mMKMnO4has an absorbance maximum of 0.26 at 525 nm in a

1.000-cm cell. Find the molarabsorptiveand the concentration of a solution whose

absorbance is 0.52 at 525 nm in the same cell.

Short Answer

Expert verified

The value of molar absorptive is2600M-1cm-1

The value of solution absorbance is0.2mM

Step by step solution

01

Calculate molar absorptive:

Equation to calculate ε is,

ε=Abc →(1)

Given,

c=0.10mMb=1.000cmA=0.26

Substitute in equation (1),

ε=0.261.000cm0.10mM10-3M1mMε=2600M-1cm-1

The value of molar absorptive is 2600M-1cm-1

02

Calculate solution concentration:

Equation for concentration is c=Abc

Given,

b=1.000cmA=0.26ε=2600M-1cm-1

Substitute in above equation,

c=0.521.000cm2600M-1cm-1=0.0002M103mM1Mc=0.2mM

The value of solution absorbance is0.2mM

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 ÷ÈÓ°Ö±²¥!

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

(a) How much energy (in kilojoules) is carried by one mole of photons of red light withλ= 650 nm?

(b) How many kilojoules are carried by one mole of photons of violet light withλ= 400 nm?

If a sample for spectrophotometric analysis is placed in a 10 - cmcell, the absorbance will be 10times greater than the absorbance in a 1 - cmcell. Will the absorbance of the reagent-blank solution also be increased by a factor of 10?

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 A1.00 - mLaliquot was withdrawn, placed in a 10 - mLvolumetric flask, and diluted to the mark.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?

Preparing standards for a calibration curve.

(a) How much ferrous ethylene diammonium sulfate(FeH3NCH2CH2NH3SO42⋅4H2O,FM382.15)should be dissolved in a 500mL volumetric flask with1MH2SO4to obtain a stock solution with~500μgFe/mL2?

(b) When making stock solution (a), you actually weighed out 1.627 g of reagent. What is the Fe concentration in role="math" localid="1668357579931" 500μgFe/mL2?

(c) How would you prepare 500 mL of standard containing 1,2,3,4, and5μgFe/mL2in0.1MH2SO4infrom stock solution (b) using any Class A pipets from Table 2-4 with only 500 -mL volumetric flasks?

(d) To reduce the generation of chemical waste, describe how you could prepare 50 mL of standard containing, andinfrom stock solution (b) by serial dilution using any Class A pipets from Table 2-3 with only 50mL volumetric flasks?

Preparing standards for a calibration curve.

(a) How much ferrous ammonium sulfate(FeNH42SO42⋅6H2OFM392.15)should be dissolved in a 500mL volumetric flask withto obtain a stock solution with1000μgFe/mL?

(b) When making stock solution (a), you weighed out 3.627 g of reagent. What is the Fe concentration in?

(c) How would you prepare 250mL of standard containing containing ~1,2,3,4,5,6,7,8and10in0.1MH2SO4infrom stock solution (b) using only 5- and 10-mL Class A pipets, only 250mL volumetric flasks, and only two consecutive dilutions of the stock solution? For example, to prepare a solution with ~4μgFe/mL,, you could first dilute 15mL(=10+5mL) of stock solution up to 250 mL mLto get~(1525)(1000μgFFe/mL)=∼60μgFe/mL Then dilute 15mL of the new solution up to 250 mL again to get~(15250)(60μgFe/mL)=∼3.6μgFe/mL

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.