Calculate the mass of ki in grams required to prepare 9.00 ✕ 102 ml of a 1.40 m solution

Volume in liters:

9.00x10² mL / 1000 => 0.9 L 

Molar mass KI => 166.0028 g/mol

Mass KI = volume x molar mass x molarity

Mass KI = 0.9 x 166.0028 x 1.40

= 209.16 g of KI

hope this helps!

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj