If 0.2997 g of potassium iodate (KIO3) are dissolved in 25.00 mL of distilled water, what is the resulting molarity (M)? Give a brief explanation about why this is a very precise concentration to use. In other words, what makes potassium iodate such a good reagent with which to work in lab?
firstly, we calculate the molarity of potassium iodate.
since , Molarity = no. Of mole/ volume (in Litre)
So, number of moles of potassium iodate= weight in g / molar mass.
= 0.2997/214 = 0.0014
And volume = 25 ml/1000l = 0.025 Litre.
Therefore, Molarity M= 0.0014/0.025 = 0.056 mol/litre.
Now, Why makes potassium iodate such a good reagent ??
the answer is -
Since potassium iodate is a salt of iodine , and it is quite stable reagent.it has much higher shelf life. It can be stored in lab and can be used for a long time.
As well, potassium iodate is a thyroid blocking agent and also it can be used as a maturing agent in baking sometimes.
These unique properties makes potassium iodate a good reagent with which to work in lab.
please upvote
If 0.2997 g of potassium iodate (KIO3) are dissolved in 25.00 mL of distilled water, what...
0.6573 g of potassium iodate is dissolved and diluted to a volume of 100.0 mL. Then 25.00 mL of this stock solution is pipetted into a 100.0 mL volumetric flask which is filled to the mark with water. What is the concentration of the final standard solution in mol/L? Include units in your answer, and round the final answer to the correct number of significant figures.
A 0.1358-g portion of potassium iodate (MW 214.00), about 2 g of potassium iodide, and 2 mL of 6 M hydrochloric acid were dissolved in 25 mL of distilled water. The triiodide formed during the ensuring reaction was titrated to the starch endpoint with 31.94 mL of a thiosulfate solution. A 25.00-mL triiodide sample solution was titrated to the endpoint with 21.33 mL of the standardized thiosulfate solution. Calculate the concentrations of the thiosulfate solution and the triiodide solution.
25.00 g of potassium hydrogen phosphate (K HPO4) is dissolved in enough water to make 125.0 mL of a solution with a density of 1.22 g/mL. What is the concentration of the solution in % (m/m). % (m/v). molarity, mosM and mEq/L of the potassium ion? Note: K2HPO4 has a molar mass of 174.2g/mole. What would be the final molarity of the above solution if you were to dilute it by adding 175 mL of water?
(a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500 mL volumetric flask. Then 50.00 mL of the solution was pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many millimoles of I3− are created by the reaction? (b) The triiodide from part (a) reacted with 37.54 mL of Na2S2O3 solution. What is the concentration of the Na2S2O3 solution? (c) A 1.223...
(a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500 mL volumetric flask. Then 50.00 mL of the solution was pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many millimoles of I3− are created by the reaction? ANSWER is 1.433 mmol (b) The triiodide from part (a) reacted with 37.54 mL of Na2S2O3 solution. What is the concentration of the Na2S2O3solution?
Exactly 0.5458 g of potassium chloride was placed in a 200 mL volumetric flask (class A). It was dissolved and diluted with distilled water up to the 200-mL mark. Calculate: (a) concentration of KCl in the resulting solution in g/L (b) concentration of potassium K in this solution in g/L (c) molarity of this solution
7.920 g of potassium oxalate is dissolved in water to make 450.0 mL of solution. What is the molarity of the potassium oxalate solution ?
15 g of potassium chlorate are dissolved in water to produce 250 mL of solution. What is the solution's molarity?
A 0.350-g sample pf an acid, HX, is dissolved in 25.00 mL of water. The resulting solution is titrated with 0.140 M NaOH solution, and 25.40 mL of the base solution is required to neutralize the acid. Calculate the molar mass of the acid.
6: Suppose 50.0 g of pure potassium hydroxide is dissolved in 250.00 mL water ( Solvent: -0.9987 g/mL). Given the density of the solution is 1.020 g/mL, what will be the molarity of the solution. (Hint: You need to find the volume of the solution)