
Normal human blood sugar range is $65 - 105mg/dL$. Considering density of human blood is $1.06kg/L$, if a patient’s sugar level reads $720ppm$, his/her blood sugar at that time is
A.Normal
B.High
C.Low
D.Can not say
Answer
515.1k+ views
Hint:We have to first calculate the mass of sugar. From the mass of sugar, we have to determine the concentration of sugar. The calculated concentration of sugar is converted into $mg/dL$ using the conversion factor. From these data, we can determine the blood sugar range of that person.
Complete step by step answer:
Given data contains,
Normal human blood sugar range is $65 - 105mg/dL$.
Density of blood is $1.06kg/L$.
Sugar level of that patient is $720ppm$.
One Litre of blood contains $1.06kg$.
We have to calculate the mass of sugar using the ppm and mass of blood.
${\text{ppm}} = \dfrac{{{\text{Mass of sugar}}}}{{{\text{Mass of blood}}}}$
On rearranging the above equation we get,
${\text{Mass of sugar}} = {\text{ppm}} \times {\text{Mass of blood}}$
Let us now substitute the values of ppm and mass of blood.
${\text{Mass of sugar}} = {\text{ppm}} \times {\text{Mass of blood}}$
Mass of sugar$ = 720 \times 1.06$
Mass of sugar$ = 763.2mg$
The mass of sugar in milligrams is $763.2mg$.
The mass of sugar in milligrams is converted into grams by dividing it by 1000.
$g = 763.2mg \times \dfrac{{1g}}{{1000mg}}$
$g = 0.763g$
The mass of sugar in grams is $0.763g$.
Let us now calculate the concentration of sugar. We know that concentration is the number of moles of solute to the volume of the solution. Here, the solute is sugar.
So, the concentration of sugar is calculated as,
${\text{Concentration = }}\dfrac{{{\text{Moles}}}}{{{\text{Volume}}}}$
We know that molar mass of sugar is $180g/mol$.
Concentration\[ = \dfrac{{\dfrac{{0.763}}{{180}}}}{1}\]
Concentration$ = 4.24mmol/L$
The concentration of sugar is $4.24mmol/L$
We know that $1mmol/L = 18mg/dL$
Therefore, $4.24mmol/L$ in $mg/dL$ is,
$mg/dL = 4.24mmol/L \times \dfrac{{18mg/dL}}{{1mmol/L}}$
$mg/dL = 76.32mg/dL$
The concentration of sugar in mg/dL is $76.32mg/dL.$
Human blood sugar is normal at that time. Normal human blood sugar lies in the range of $65 - 105mg/dL$. Considering the density of human blood as $1.06kg/L$, a patient's sugar level reads $720ppm$ thus his/her blood sugar at that time is normal.
Therefore, the option (A) is correct.
Note:
If the sugar level of the blood lies in the value of $7.8mmol/L$, then the level of sugar is normal. If the level of sugar in blood lies more than $11.1mmol/L$, then the condition is said to be diabetes, and if sugar level lies between $7.8mmol/L$ and $11.1mmol/L$, then the condition is said to be prediabetes.
Complete step by step answer:
Given data contains,
Normal human blood sugar range is $65 - 105mg/dL$.
Density of blood is $1.06kg/L$.
Sugar level of that patient is $720ppm$.
One Litre of blood contains $1.06kg$.
We have to calculate the mass of sugar using the ppm and mass of blood.
${\text{ppm}} = \dfrac{{{\text{Mass of sugar}}}}{{{\text{Mass of blood}}}}$
On rearranging the above equation we get,
${\text{Mass of sugar}} = {\text{ppm}} \times {\text{Mass of blood}}$
Let us now substitute the values of ppm and mass of blood.
${\text{Mass of sugar}} = {\text{ppm}} \times {\text{Mass of blood}}$
Mass of sugar$ = 720 \times 1.06$
Mass of sugar$ = 763.2mg$
The mass of sugar in milligrams is $763.2mg$.
The mass of sugar in milligrams is converted into grams by dividing it by 1000.
$g = 763.2mg \times \dfrac{{1g}}{{1000mg}}$
$g = 0.763g$
The mass of sugar in grams is $0.763g$.
Let us now calculate the concentration of sugar. We know that concentration is the number of moles of solute to the volume of the solution. Here, the solute is sugar.
So, the concentration of sugar is calculated as,
${\text{Concentration = }}\dfrac{{{\text{Moles}}}}{{{\text{Volume}}}}$
We know that molar mass of sugar is $180g/mol$.
Concentration\[ = \dfrac{{\dfrac{{0.763}}{{180}}}}{1}\]
Concentration$ = 4.24mmol/L$
The concentration of sugar is $4.24mmol/L$
We know that $1mmol/L = 18mg/dL$
Therefore, $4.24mmol/L$ in $mg/dL$ is,
$mg/dL = 4.24mmol/L \times \dfrac{{18mg/dL}}{{1mmol/L}}$
$mg/dL = 76.32mg/dL$
The concentration of sugar in mg/dL is $76.32mg/dL.$
Human blood sugar is normal at that time. Normal human blood sugar lies in the range of $65 - 105mg/dL$. Considering the density of human blood as $1.06kg/L$, a patient's sugar level reads $720ppm$ thus his/her blood sugar at that time is normal.
Therefore, the option (A) is correct.
Note:
If the sugar level of the blood lies in the value of $7.8mmol/L$, then the level of sugar is normal. If the level of sugar in blood lies more than $11.1mmol/L$, then the condition is said to be diabetes, and if sugar level lies between $7.8mmol/L$ and $11.1mmol/L$, then the condition is said to be prediabetes.
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