Question

How many moles of solute particles are present in 4.98 mL of 0.72 M (NH4)2SO4? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.

Answer 1

First, calculate the moles of (NH4)2SO4 using the molarity and volume, then find the base 10 logarithm of the result. The moles of (NH4)2SO4 are approximately 0.003586, and the logarithm is roughly -2.45.

Step-by-step explanation:

To compute the moles of solute in a given volume of solution, one can use the formula moles of solute = Molarity (mol/L) × Volume of solution (L). For a 0.72 M solution of (NH4)2SO4, when given a volume of 4.98 mL (which is 0.00498 L), we calculate:

moles of solute = 0.72 mol/L × 0.00498 L = 0.0035856 mol

Now, to find the base 10 logarithm, we perform the following calculation:

LOG(moles of solute) = LOG(0.0035856) ≈ -2.45

Answer 2

Answer: The number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}`

Molarity of ammonium sulfate solution = 0.72 M

Volume of solution = 4.98 mL

Putting values in above equation, we get:

`0.72M=\frac{\text{Moles of ammonium sulfate}* 1000}{4.98mL}\\\\\text{Moles of ammonium sulfate}=0.0036mol`

Taking the log (base 10) of the calculated moles of ammonium sulfate we get:

`\log_(10)(0.0036)=-2.44`

Hence, the number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44