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* 10^(-10)M`

Step-by-step 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}* \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}=(0.000,000,0342g)/(329.21g/mol* 1L)\\\\\text{Molarity of solution}=1.0388* 10^(-10)M`

Hence, the molarity of solution is
`1.0388* 10^(-10)M`