Achemist prepares a solution of potassium permanganate (kmno4) by measuring out 3.8 umol of potassium permanganate into a 100 ml volumetric flask and filling the flask to the mark with water. calculate the concentration in mol/l of the chemist's potassium permanganate solution. round your answer to 2 significant digits. x 5 ? explanation check

3,8×10⁻⁵ mol/L of potassium permanganate solution

Explanation:

To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.

3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.

100 mL × ( 1L / 1000 mL) = 0,100 L

Thus, concentration in mol/L is:

3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution

I hope it helps!

what are the choices?

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm