How many moles of ammonium nitrate (NH4NO3) are present in 0.085 L of a 0.23 M ammonium nitrate (NH4NO3) solution?

Respuesta :

Answer:

No. of moles [tex]\approx[/tex] 0.020

Explanation:

To calculate the number of moles of a substance in a solution given its concentration and volume, we use the following formula:

[tex]\boxed{\mathrm{No. \ of \ moles = Concentration \times Volume}}[/tex],

where concentration is in mol/dm³ (or M) and volume is in L.

In the question, we are told that the volume of the solution is 0.085 L and that its concentration is 0.23 M. Substituting these values into the formula above, we get:

No. of moles = 0.23 × 0.085

                     = 0.01955

                     [tex]\approx[/tex] 0.020 mol  (3 s.f.)

Therefore, there are 0.020 moles of NH₄NO₃ in the solution.