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Answer :
Final Answer:
B) 75 mEq sodium are contained in 250 ml of a solution of sodium bicarbonate that is isoosmotic with plasma that is 300 mosml. Thus option B is correct
Explanation:
To determine the milliequivalents (mEq) of sodium in the solution of sodium bicarbonate, we can use the formula: mEq = concentration (in mmolL) × volume (in L).
Given that the solution is isoosmotic with plasma at 300 mosmL and assuming sodium bicarbonate dissociates into sodium ions (Na⁺) and bicarbonate ions (HCO₃⁻), we know that each molecule of sodium bicarbonate contributes one sodium ion. Therefore, the concentration of sodium ions in the solution is also 300 mmolL.
Converting the volume of the solution to liters (250 ml = 0.25 L), we can calculate the milliequivalents of sodium: mEq = 300 mmolL × 0.25 L = 75 mEq.
The solution's isoosmotic nature ensures that its total solute concentration matches that of plasma, which is approximately 300 mosmL.
Since sodium ions contribute significantly to the osmolarity of the solution and each sodium bicarbonate molecule yields one sodium ion upon dissociation, the concentration of sodium ions in the solution mirrors the osmolarity. Therefore, the concentration of sodium ions is also 300 mmolL.
By multiplying the concentration of sodium ions by the volume of the solution in liters, we can find the total milliequivalents of sodium present. In this case, the volume is 250 ml, which is equivalent to 0.25 liters.
Thus, the calculation becomes straightforward: mEq = 300 mmolL × 0.25 L = 75 mEq. Therefore, the solution contains 75 milliequivalents of sodium.
Therefore option B is correct.
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