To make a 10% KI solution, dissolve 10 grams of potassium iodide (KI) in 90 grams of water, for a total of 100 grams of solution. This will give you a 10% weight/volume (w/v) solution of KI.
Molarity = moles of solute/Liters of solution. get moles KI 2.822 grams KI (1 mole KI/166 grams) = 0.017 moles KI ( 67.94 ml = 0.06794 Liters ) Molarity = 0.017 moles KI/0.06794 Liters = 0.2502 M KI
To calculate the mass of KI in the solution, first calculate the number of moles of KI present using the formula moles = Molarity x Volume (in liters). Then, use the molar mass of KI (potassium iodide) to convert moles to grams. The molar mass of KI is 166 g/mol.
Hi, having a solution of KI implies having H20 so tin iodide will hydrolyze forming the tin oxide and HI. Another interesting reaction is when you add KI (in solid phase) into tin iodide solution (dissolved in acetone). You will obtain a salt like this: K(SnI5) I hope I helped you. Note that this answer could be wrong, I'm currently a student of chemistry and I have a lot to learn yet.
To find the molarity (M) of the KI solution, first convert grams of KI to moles using its molar mass (KI's molar mass is approximately 166 g/mol). Moles of KI = 36.52 g / 166 g/mol ≈ 0.220 moles. Then, convert the volume from mL to liters: 820 mL = 0.820 L. Finally, calculate the molarity: M = moles/volume = 0.220 moles / 0.820 L ≈ 0.268 M. Thus, the molarity of the KI solution is approximately 0.268 M.
The answer is 0,1 mol.
0.18M
To prepare a 10% potassium iodide solution, dissolve 10 grams of potassium iodide in 90 ml of water. Don't forget to wear appropriate protective gear like gloves and goggles. Stir the mixture well until the potassium iodide is fully dissolved.
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
Molarity = moles of solute/Liters of solution. get moles KI 2.822 grams KI (1 mole KI/166 grams) = 0.017 moles KI ( 67.94 ml = 0.06794 Liters ) Molarity = 0.017 moles KI/0.06794 Liters = 0.2502 M KI
To prepare iodine solution, dissolve iodine crystals in a mixture of water and potassium iodide (KI). The ratio of iodine to KI will determine the concentration of the solution. The solution should be stored in a dark bottle to prevent degradation from light exposure.
Potassium iodide (KI) is considered a neutral solution because it dissociates in water to form potassium ions (K^+) and iodide ions (I^-), which do not affect the pH of the solution. The ions from KI do not contribute to the acidity or basicity of the solution, meaning that KI does not alter the pH level significantly.
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
To calculate the mass of KI in the solution, first calculate the number of moles of KI present using the formula moles = Molarity x Volume (in liters). Then, use the molar mass of KI (potassium iodide) to convert moles to grams. The molar mass of KI is 166 g/mol.
KI (potassium iodide) is a salt that dissociates into K+ and I- ions in water. Both potassium ions and iodide ions are neutral and do not affect the pH of a solution. Therefore, the pH of a solution of KI would remain unchanged.
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The percent composition of potassium in potassium iodide (KI) is 58.5%. This is calculated by dividing the atomic mass of potassium by the molar mass of KI and multiplying by 100.