/*! 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} Problem 88 Fish-liver oil is a good source ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 88

Fish-liver oil is a good source of vitamin \(\mathrm{A},\) whose concentration is measured spectrometrically at a wavelength of \(329 \mathrm{nm}\) (a) Suggest a reason for using this wavelength. (b) In what region of the spectrum does this wavelength lie? (c) When \(0.1232 \mathrm{~g}\) of fish-liver oil is dissolved in \(500 . \mathrm{mL}\) of solvent, the absorbance is 0.724 units. When \(1.67 \times 10^{-3} \mathrm{~g}\) of vitamin \(\mathrm{A}\) is dissolved in \(250 . \mathrm{mL}\) of solvent, the absorbance is 1.018 units. Calculate the vitamin A concentration in the fish-liver oil.

Short Answer

Expert verified
The wavelength 329 nm is used because it is the peak absorbance of vitamin A, which lies in the UV region. The concentration of vitamin A in the fish-liver oil is 0.0193 g/g.

Step by step solution

01

Understanding the use of 329 nm wavelength

The wavelength of 329 nm is chosen because vitamin A has its maximum absorbance (peak absorbance) at this wavelength. This ensures that the measurement is accurate and specific to vitamin A, minimizing the interference from other substances.
02

Spectrum Region Identification

The wavelength of 329 nm falls within the ultraviolet (UV) region of the electromagnetic spectrum, which ranges from approximately 10 nm to 400 nm.
03

Calculate the Molar Absorptivity (蓻) for Vitamin A

Using the Beer-Lambert Law: \[ A = 蓻 \times c \times l \]Rearrange to solve for molar absorptivity (蓻): \[ 蓻 = \frac{A}{c \times l} \]Given: \[ A = 1.018 \]\[ c = \frac{1.67 \times 10^{-3} \text{ g}}{250 \text{ mL}} \text{ Vitamin A} \] (we assume l is 1 cm) Convert concentration to g/mL: \[ c = \frac{1.67 \times 10^{-3}}{0.25} = 6.68 \times 10^{-3} \text{ g/mL} \]So, \[ 蓻 = \frac{1.018}{(6.68 \times 10^{-3} \text{ g/mL}) \times 1 \text{ cm}} \]\[ 蓻 = 152.4 \text{ mL/(g cm)} \]
04

Calculate the Concentration of Vitamin A in Fish-Liver Oil

Again using Beer-Lambert Law: \[ c = \frac{A}{蓻 \times l} \]Given: \[ A = 0.724 \]Using the molar absorptivity calculated, and assuming path length l is 1 cm: \[ c = \frac{0.724}{152.4 \times 1} = 4.75 \times 10^{-3} \text{ g/mL} \]This is the concentration in the solution. Now convert this to the concentration in the fish-liver oil: \[ 4.75 \times 10^{-3} \text{ g/mL} \times 500 \text{ mL} = 2.375 \text{ g} \]\[ \text{Concentration of Vitamin A in the fish-liver oil} = \frac{2.375 \text{ g}}{0.1232 \text{ g}} = 0.0193 \text{ g/g} \]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

spectrophotometry
Spectrophotometry is a technique used to measure the amount of light absorbed by a solution. This helps in determining the concentration of a substance in that solution. The core principle is the Beer-Lambert Law, which relates the absorbance with concentration using the formula: \( A = 蓻 \times c \times l \).
Here:
  • \

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

Compare the wavelengths of an electron (mass \(=9.11 \times 10^{-31} \mathrm{~kg}\) ) and a proton (mass \(=1.67 \times 10^{-27} \mathrm{~kg}\) ), each having (a) a speed of \(3.4 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a kinetic energy of \(2.7 \times 10^{-15} \mathrm{~J}\)

If particles have wavelike motion, why don't we observe that motion in the macroscopic world?

One reason carbon monoxide (CO) is toxic is that it binds to the blood protein hemoglobin more strongly than oxygen does. The bond between hemoglobin and CO absorbs radiation of \(1953 \mathrm{~cm}^{-1}\). (The unit is the reciprocal of the wavelength in centimeters.) Calculate the wavelength (in \(\mathrm{nm}\) and \(\dot{\mathrm{A}}\) ) and the frequency (in \(\mathrm{Hz}\) ) of the absorbed radiation.

Electric power is measured in watts \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) .\) About \(95 \%\) of the power output of an incandescent bulb is converted to heat and \(5 \%\) to light. If \(10 \%\) of that light shines on your chemistry textbook, how many photons per second shine on the book from a 75-W bulb? (Assume that the photons have a wavelength of \(550 \mathrm{nm} .\) )

Only certain transitions are allowed from one energy level to another. In one- electron species, the change in \(I\) for an allowed transition is \(\pm 1 .\) For example, a \(3 p\) electron can move to a \(2 s\) orbital but not to a \(2 p\). Thus, in the UV series, where \(n_{\text {final }}=1\), allowed transitions can start in a \(p\) orbital \((l=1)\) of \(n=2\) or higher, not in an \(s(l=0)\) or \(d(l=2)\) orbital of \(n=2\) or higher. From what orbital do each of the allowed transitions start for the first four emission lines in the visible series \(\left(n_{\text {final }}=2\right) ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

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.