Question

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

Answer 1

Answer: The molarity of solution is $1.0388\times 10^{-10}M$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Mass of solute (CAMP) = 0.0000000342 g

Molar mass of CAMP = 329.21 g/mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{0.000,000,0342g}{329.21g/mol\times 1L}\\\\\text{Molarity of solution}=1.0388\times 10^{-10}M$

Hence, the molarity of solution is $1.0388\times 10^{-10}M$