
Calculate the percentage by mass to volume for $0.2M\,NaOH$ solution.
Answer
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Hint:
$0.2M$ means that $0.2$ moles in $1L$ of the solution.
$number\,of\,moles = \dfrac{{Given\,mass}}{{molecular\,mass}}$
The molecular mass of $NaOH = 40g$
Hence, use this to find the value of given mass.
This value of given mass will be present in $1L$ of the solution.
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
Substitute the value of given mass from above equation and volume= $1L$
Formula Used:
$mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
Complete answer:
Mass by Volume percentage signifies how much amount of solute is present in how much amount of the solution.
It can represented mathematically as:
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
In the question given to us, we have the value of the molarity of the solution which is $0.2M\,NaOH$.
The formula for molarity is:
$Molarity = \dfrac{{number\,of\,moles\,of\,solute}}{{volume\,of\,solution\,(in\,Liter)}}$
If we were to represent this formula in word format, it would be the number of moles of solute present per litre of the solution.
We know that,
$Number\,of\,moles = \dfrac{{Given\,mass}}{{molar\,mass}}$
Molar mass of $NaOH = Mass\,of\,Na + Mass\,of\,O + Mass\,of\,H$
$Mass\,of\,Na = 23g,Mass\,of\,O = 16g,Mass\,of\,H = 1g$
Substituting the above values to get the molecular mass of Sodium Hydroxide;
We get;
$Mass\,of\,NaOH = 23 + 16 + 1 \Rightarrow 40g$
From the information given in the question we know molarity of $NaOH$ is $0.2$ .
Hence there are $0.2$ moles of $NaOH$ in $1L$ of the solution
Substituting the value of molar mass of $NaOH$ and the number of moles in the formula
$Number\,of\,moles = \dfrac{{Given\,mass}}{{molar\,mass}}$
we get;
$0.2 = \dfrac{{given\,mass}}{{40g}}$
Solving for the value of given mass we get,
$given\,mass = 8g$ .
Hence, $8g$ of $NaOH$ is present in $1L$ of the solution.
$\therefore Mass = 8g,Volume = 1L$
Substituting these values in the formula:
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
we get,
$mass\,percentage = \dfrac{{8g}}{{1L}} \times 100$
solving this equation we get the value of mass to volume percentage as $8\% $.
Note: Mass by Volume percentage is also known as mass concentration. It cannot be used for industrial application such as determining the concentration in medicines because volume acts as a function of temperature and as temperature changes the value of volume will also change and hence the value of mass concentration will change.
$0.2M$ means that $0.2$ moles in $1L$ of the solution.
$number\,of\,moles = \dfrac{{Given\,mass}}{{molecular\,mass}}$
The molecular mass of $NaOH = 40g$
Hence, use this to find the value of given mass.
This value of given mass will be present in $1L$ of the solution.
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
Substitute the value of given mass from above equation and volume= $1L$
Formula Used:
$mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
Complete answer:
Mass by Volume percentage signifies how much amount of solute is present in how much amount of the solution.
It can represented mathematically as:
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
In the question given to us, we have the value of the molarity of the solution which is $0.2M\,NaOH$.
The formula for molarity is:
$Molarity = \dfrac{{number\,of\,moles\,of\,solute}}{{volume\,of\,solution\,(in\,Liter)}}$
If we were to represent this formula in word format, it would be the number of moles of solute present per litre of the solution.
We know that,
$Number\,of\,moles = \dfrac{{Given\,mass}}{{molar\,mass}}$
Molar mass of $NaOH = Mass\,of\,Na + Mass\,of\,O + Mass\,of\,H$
$Mass\,of\,Na = 23g,Mass\,of\,O = 16g,Mass\,of\,H = 1g$
Substituting the above values to get the molecular mass of Sodium Hydroxide;
We get;
$Mass\,of\,NaOH = 23 + 16 + 1 \Rightarrow 40g$
From the information given in the question we know molarity of $NaOH$ is $0.2$ .
Hence there are $0.2$ moles of $NaOH$ in $1L$ of the solution
Substituting the value of molar mass of $NaOH$ and the number of moles in the formula
$Number\,of\,moles = \dfrac{{Given\,mass}}{{molar\,mass}}$
we get;
$0.2 = \dfrac{{given\,mass}}{{40g}}$
Solving for the value of given mass we get,
$given\,mass = 8g$ .
Hence, $8g$ of $NaOH$ is present in $1L$ of the solution.
$\therefore Mass = 8g,Volume = 1L$
Substituting these values in the formula:
$\therefore mass\,by\,volume\,percentage = \dfrac{{mass}}{{volume}} \times 100$
we get,
$mass\,percentage = \dfrac{{8g}}{{1L}} \times 100$
solving this equation we get the value of mass to volume percentage as $8\% $.
Note: Mass by Volume percentage is also known as mass concentration. It cannot be used for industrial application such as determining the concentration in medicines because volume acts as a function of temperature and as temperature changes the value of volume will also change and hence the value of mass concentration will change.
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