/*! 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 27 \(10.30 \mathrm{mg}\) of \(\math... [FREE SOLUTION] | 91Ó°ÊÓ

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\(10.30 \mathrm{mg}\) of \(\mathrm{O}_{2}\) is dissolved into a liter of sea water of density \(1.03\) \(\mathrm{g} / \mathrm{mL}\). The concentration of \(\mathrm{O}_{2}\) in ppm is Main Jan. 09. 2020 (II)

Short Answer

Expert verified
The concentration of \(\mathrm{O}_{2}\) in sea water is 10.01 ppm.

Step by step solution

01

Convert Milligrams to Grams

First, we must convert the mass of \(\mathrm{O}_{2}\) from milligrams to grams for consistency in units. Since there are 1000 milligrams in a gram, we have: \ 10.30 \, \mathrm{mg} = \frac{10.30}{1000} \, \mathrm{g} = 0.0103 \, \mathrm{g}\.
02

Calculate the Density of Sea Water in Grams per Liter

Given that the density of sea water is \(1.03 \, \mathrm{g/mL}\), we convert this to \(\mathrm{g/L}\). Since there are 1000 mL in a liter, the density of sea water is: \ 1.03 \, \mathrm{g/mL} \times 1000 \, \mathrm{mL/L} = 1030 \, \mathrm{g/L}\.
03

Calculate the Fraction of \\(\mathrm{O}_{2}\\) by Mass

To find the fraction by mass of \(\mathrm{O}_{2}\), divide the mass of \(\mathrm{O}_{2}\) by the mass of one liter of sea water: \ \frac{0.0103 \, \mathrm{g}}{1030 \, \mathrm{g}} = 1.00097 \times 10^{-5}\.
04

Convert the Fraction to Parts Per Million (ppm)

Parts per million is defined as parts of solute per million parts of solution by mass. Thus, multiply the fraction of \(\mathrm{O}_{2}\) by one million: \ 1.00097 \times 10^{-5} \times 10^6 = 10.0097 \, \text{ppm}\. Round this to four significant figures to match the given data: \ \text{10.01} \, \text{ppm}\.

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Key Concepts

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

Oxygen Concentration
Oxygen concentration in water, such as seawater, is an essential metric to understand. For marine life to thrive, the right levels of dissolved oxygen must be maintained. In this particular problem, we're interested in expressing the concentration of oxygen in a very specific unit called parts per million (ppm).

In chemistry, ppm is used to measure the small concentrations of solutes in a solution. It refers to the number of parts of a substance in one million parts of the total solution. Calculating the ppm involves converting the mass of the solute (oxygen in our case) into a comparable unit with the density of the solution, eventually multiplying by one million to find the desired concentration value. The challenge lies in ensuring all units are consistent before performing these conversions.
Density Conversion
Density is the mass of a substance per unit volume. For scenarios like this problem where we're dealing with solutions and small concentrations, understanding the density in correct units is crucial. Here, we start with the density of seawater given in grams per milliliter (g/mL).

To efficiently utilize this information in calculations with larger units such as liters, conversion is necessary. Understanding that there are 1000 milliliters in a liter allows you to switch from g/mL to g/L by multiplying the density by 1000. Similarly, this fundamental conversion step ensures that calculations involving mass of the solute and the solution volume match in terms of units, easing further calculations.
Mass Fraction
A mass fraction describes how much of one substance is in a mixture, expressed as a fraction of the total mass. It is an important concept, especially in diluted solutions, to accurately determine the concentration of a specific component.

In the context of this problem, the mass fraction of oxygen is found by dividing the mass of oxygen by the total mass of seawater per liter. While it seems simple, it effectively scales down the larger mass of the solvent (seawater) against the relatively diminutive mass of the oxygen solute, creating a dimensionless number that provides insight into the oxygen's relative abundance in the mixture. This mass fraction is the stepping stone to converting the values into ppm, a more practical unit for parts-per notation.
Unit Conversion
Unit conversion is a cornerstone in solving chemistry problems involving different measurements. It ensures that you are working with comparable units, thus maintaining the integrity of your calculation across various dimensions.

This involves:
  • Converting milligrams to grams for ease of mass comparisons.
  • Switching milliliters to liters to express density for volumetric computations.
  • Scaling fractions by a factor (such as one million) to shift into parts per notation (ppm).
Accuracy in unit conversion prevents errors at each solution step. It ties together concepts like mass fraction and density conversion cohesively, ultimately leading to the correct determination of concentration expressed in ppm.

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Most popular questions from this chapter

(a) One litre of a sample of hard water contains \(1 \mathrm{mg}\) of \(\mathrm{CaCl}_{2}\) and 1 \(\mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). Find the total hardness in terms of parts of \(\mathrm{CaCO}_{3}\) per \(10^{6}\) parts of water by weight. (b) A sample of hard water contains \(20 \mathrm{mg}\) of \(\mathrm{Ca}^{++}\)ions per litre. How many milli-equivalent of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) would be required to soften 1 litre of the sample? (c) \(1 \mathrm{~g}\) of \(\mathrm{Mg}\) is burnt in a closed vessel which contains \(0.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of \(0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) will dissolve the residue in the vessel.

(i) A sample of \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is strongly heated in air. The residue is \(\mathrm{Mn}_{3} \mathrm{O}_{4}\)(ii) The residue is dissolved in \(100 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{FeSO}_{4}\) containing dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (iii) The solution reacts completely with \(50 \mathrm{~mL}\) of \(\mathrm{KMnO}_{4}\) solution. (iv) \(25 \mathrm{~mL}\) of the \(\mathrm{KMnO}_{4}\) solution used in step (iii) requires \(30 \mathrm{~mL}\) of \(0.1\) N \(\mathrm{FeSO}_{4}\) solution for complete reaction. Find the amount of \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) present in the sample. [1980]

The largest number of molecules is in (a) \(36 \mathrm{~g}\) of water (b) \(28 \mathrm{~g}\) of carbon monoxide (c) \(46 \mathrm{~g}\) of ethyl alcohol (d) \(54 \mathrm{~g}\) of nitrogen pentoxide

An aqueous solution containing \(0.10 \mathrm{~g} \mathrm{KIO}_{3}\) (formula weight \(=214.0\) ) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated \(\mathrm{I}_{2}\) consumed \(45.0 \mathrm{~mL}\) of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution. [1998-5 Marks]

Mixture \(X=0.02 \mathrm{~mol}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02 \mathrm{~mol} \mathrm{o}\) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution. [2003S] 1 litre of mixture \(X+\) excess \(\mathrm{AgNO}_{3} \rightarrow Y\) 1 litre of mixture \(X+\) excess \(\mathrm{BaCl}_{2} \rightarrow Z\) No. of moles of \(Y\) and \(Z\) are (a) \(0.01,0.01\) (b) \(0.02,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

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