A chemist prepares a solution of potassium iodide (KI) by measuring out 88. umol of potassium iodide into a 200. mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's potassium iodide solution. Round your answer to 2 significant digits.

Please Help me!

The concentration in mol/L of the chemist's potassium iodide solution : = 4.4 x 10⁻⁴ M

Given

88. μmol of Potassium iodide

200. mL volumetric flask

Required

The concentration

Solution

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Conversion :

88. μmol = 88 . 10⁺⁶ mol = 8.8 10⁻⁵ mol

200 ml = 0.2 L

The molarity :

= n : V

= 8.8.10⁻⁵ : 0.2

= 4.4 x 10⁻⁴ M

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2