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Question: Average human blood serum contains 18 mg of \({{K}^{+}}\) and 365 mg of \(C{{l}^{-}}\) per 100 L. Th...

Average human blood serum contains 18 mg of K+{{K}^{+}} and 365 mg of ClC{{l}^{-}} per 100 L. The molar concentration for  Cl~C{{l}^{-}} ion is y×103y\times {{10}^{-3}} M, (density of the serum = 1.00g/L), then y is ---------.

Explanation

Solution

There is a formula to calculate the molarity of the solution and it is as follows.
M=WMW×1V(liters)M=\dfrac{W}{MW}\times \dfrac{1}{V(liters)}
Where M = Molarity of the solution
MW = Molecular weight of the solute
W = Weight of the solute
V = Volume in liters

Complete step by step answer:
- In the question it is given that 100 ml of human serum contains 18 mg of K+{{K}^{+}} and 365 mg of ClC{{l}^{-}}.
- It is also given that to find the molar concentration of  Cl~C{{l}^{-}} .
- To calculate the molar concentration of  Cl~C{{l}^{-}} , we have to convert the 100 ml of 365 mg of  Cl~C{{l}^{-}} into liters.
- Means 1 liter of 1000 ml of blood serum contains 3650 mg or 3.65 grams of  Cl~C{{l}^{-}} .
- The molecular weight of chlorine is 35.5 gm.

- Substitute all the known values in the below formula to calculate the molar concentration of  Cl~C{{l}^{-}}.

& \Rightarrow M=\dfrac{W}{MW}\times \dfrac{1}{V(liters)} \\\ & \Rightarrow M=\dfrac{3.65}{35.5}\times \dfrac{1}{1} \\\ & \Rightarrow M=0.103 \\\ & \Rightarrow M=103\times {{10}^{-3}} \\\ \end{aligned}$$ \- In the question the concentration of chlorine is given that $y\times {{10}^{-3}}$ M. \- Comparing the answer we got from the above calculation with the given value, we will get y = 103. **Note:** The concentration of the chloride ion is very high (365 mg) when compared to potassium ion (18 mg) in 100 ml of the human serum. Generally the concentrations of the solutions are going to be expressed in molarity. Molarity is going to be denoted with a symbol M.