
One kilogram of seawater sample contains \[6mg\] dissolved \[{O_2}\]. The concentration \[{O_2}\] in the sample \[ppm\] is:
A. \[0.6\]
B. \[6.0\]
C. \[60.0\]
D. \[16.0\]
Answer
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Hint: To solve the above problem we will use some basic concepts of chemistry. We know that \[ppm\] represents parts per million. Parts per million \[\left( {ppm} \right)\] is one of the measures of representing concentration. Here we have one kilogram \[\left( {1Kg} \right)\] of the seawater sample. So using the basic concepts of measurements we will calculate the concentration in parts per million \[\left( {ppm} \right)\].
Formula Used: Concentration in \[ppm = \dfrac{{{W_{solute}}}}{{{W_{sample}}}} \times {10^6}\]
Where \[{W_{solute}}\] represents the amount of solute in \[\left( {mg} \right)\] and \[{W_{sample}}\] represents the mass of the sample in \[\left( {mg} \right)\].
Complete step by step answer:
First, we will understand the important measure of the concentration of parts per million \[\left( {ppm} \right)\]. Parts per million is the way of representing the dilute concentration of substances. The measure of concentration is in order of \[\left( {{{10}^6}} \right)\]. Now we will understand and extract the given quantities from the problem. We have given that One kilogram of seawater sample contains \[6mg\] dissolved \[{O_2}\]. So here we know that one kilogram of seawater sample corresponds to \[\left( {{{10}^6}mg} \right)\] of seawater sample which contains \[6mg\] dissolved\[{O_2}\].
Now here we have the given quantities as the amount of solute \[{W_{solute}} = 6mg\] and the amount of the mass of seawater sample is \[{W_{sample}} = 1kg = {10^6}mg\]. Now we have all the required quantities to calculate the concentration of \[{O_2}\] the sample \[ppm\]. So, we will substitute the values in the formula.
Concentration in \[ppm = \dfrac{{{W_{solute}}}}{{{W_{sample}}}} \times {10^6} = \dfrac{{6mg}}{{{{10}^6}mg}} \times {10^6} = 6.0\]
So, here the calculated value of concentration \[{O_2}\] in the sample \[ppm\] is \[6.0\].
So, the correct answer is Option B.
Note: The measure of concentration parts per million \[\left( {ppm} \right)\] is described as the concentration of something in water or soil. It also measures the mass of chemicals or contaminates per volume. It is equivalent to the fractional amount multiplied by one million.
Formula Used: Concentration in \[ppm = \dfrac{{{W_{solute}}}}{{{W_{sample}}}} \times {10^6}\]
Where \[{W_{solute}}\] represents the amount of solute in \[\left( {mg} \right)\] and \[{W_{sample}}\] represents the mass of the sample in \[\left( {mg} \right)\].
Complete step by step answer:
First, we will understand the important measure of the concentration of parts per million \[\left( {ppm} \right)\]. Parts per million is the way of representing the dilute concentration of substances. The measure of concentration is in order of \[\left( {{{10}^6}} \right)\]. Now we will understand and extract the given quantities from the problem. We have given that One kilogram of seawater sample contains \[6mg\] dissolved \[{O_2}\]. So here we know that one kilogram of seawater sample corresponds to \[\left( {{{10}^6}mg} \right)\] of seawater sample which contains \[6mg\] dissolved\[{O_2}\].
Now here we have the given quantities as the amount of solute \[{W_{solute}} = 6mg\] and the amount of the mass of seawater sample is \[{W_{sample}} = 1kg = {10^6}mg\]. Now we have all the required quantities to calculate the concentration of \[{O_2}\] the sample \[ppm\]. So, we will substitute the values in the formula.
Concentration in \[ppm = \dfrac{{{W_{solute}}}}{{{W_{sample}}}} \times {10^6} = \dfrac{{6mg}}{{{{10}^6}mg}} \times {10^6} = 6.0\]
So, here the calculated value of concentration \[{O_2}\] in the sample \[ppm\] is \[6.0\].
So, the correct answer is Option B.
Note: The measure of concentration parts per million \[\left( {ppm} \right)\] is described as the concentration of something in water or soil. It also measures the mass of chemicals or contaminates per volume. It is equivalent to the fractional amount multiplied by one million.
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