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How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
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
What is the molarity of 67.94 mL aqueous solution containing 2.822 g 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 225 mL of a 0.80M solution of KI a student adds enough water to make 1.0 L of a more dilute KI solution What is the molarity of the new solution?
0.18M
What mass of KI is present in 276 mL of a 0.50 M 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.
How many moles of KI are present in 500 ml of a 0.2M solution?
The answer is 0,1 mol.
Calculate the mass of ki in grams required to prepare 5.00 102 ml of a 2.80 m solution?
To calculate the mass of KI needed for a 2.80 m solution, we first need to know the molality (m) and the volume of the solution. The molality is defined as moles of solute per kilogram of solvent. For a 2.80 m solution, we have 2.80 moles of KI per kg of solvent. Given the volume of 5.00 x 10² mL (or 0.500 kg of water, assuming the density of water is 1 g/mL), we can use the formula: Mass of KI = moles of KI × molar mass of KI. Calculating, we find that 2.80 moles of KI correspond to about 2.80 × 166 g/mol (molar mass of KI) = 464.8 g of KI. Thus, you need approximately 464.8 grams of KI.
What is the molarity of a solution prepared by dissolving 2.41 g of potassium iodide KI in 100 mL of water?
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================
What is the molarity of 820mL of Kl solution that contains 36.52g of Kl?
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.
What is the molarity of 820 mL of KI solution that contains 36.52 g of KI?
To find the molarity, first calculate the number of moles of KI using its molar mass (KI has a molar mass of approximately 166 g/mol). The number of moles is calculated as follows: [ \text{Moles of KI} = \frac{36.52 \text{ g}}{166 \text{ g/mol}} \approx 0.22 \text{ moles} ] Next, convert the volume from mL to liters: ( 820 \text{ mL} = 0.820 \text{ L} ). Finally, calculate the molarity (M): [ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.22 \text{ moles}}{0.820 \text{ L}} \approx 0.27 \text{ M} ] Thus, the molarity of the KI solution is approximately 0.27 M.
HOW TO prepare 0.1 n iodide solution in 500 ml water?
0.1 N iodide would be 0.1 moles of the iodide salt (e.g. KI) per liter of solution. For 500 ml, you would need 0.05 moles of the iodide salt. You need to state the salt (KI, NaI, LiI, etc.) in order to determine the actual mass required.
How Wagner react to alkaloid?
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
What is the molarity of 820 mL of potassium iodide solution that contains 36.52 g of potassium iodide?
To calculate the molarity (M) of the potassium iodide (KI) solution, first determine the number of moles of KI. The molar mass of KI is approximately 166 g/mol, so: [ \text{Moles of KI} = \frac{36.52 , \text{g}}{166 , \text{g/mol}} \approx 0.219 , \text{mol} ] Next, convert the volume from mL to liters: [ 820 , \text{mL} = 0.820 , \text{L} ] Now, calculate the molarity: [ M = \frac{0.219 , \text{mol}}{0.820 , \text{L}} \approx 0.267 , \text{M} ] Thus, the molarity of the potassium iodide solution is approximately 0.267 M.
