What is the molarity of .0000000342 grams of adenosine 3'5' cyclic monophosphate (CAMP) in one lite

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Question:

What is the molarity of .0000000342 grams of adenosine 3'5' cyclic monophosphate (CAMP) in one lite

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Answer: The molarity of solution is [tex]1.0388\times 10^{-10}M[/tex]Explanation:To calculate the molarity of solution, we use the equation:[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]We are given:Mass of solute (CAMP) = 0.0000000342 gMolar mass of CAMP = 329.21 g/molVolume of solution = 1 LPutting values in above equation, we get:[tex]\text{Molarity of solution}=\frac{0.000,000,0342g}{329.21g/mol\times 1L}\\\\\text{Molarity of solution}=1.0388\times 10^{-10}M[/tex]Hence, the molarity of solution is [tex]1.0388\times 10^{-10}M[/tex]

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