A potassium iodide solution is made by mixing 0.520 g of potassium iodide with 120.0 mL of water. Calculate the molality of potassium iodide. Assume that the density of water is 1.000 g/mL
11. 22 g of potassium iodide is diluted in 1.5 L of sterile water for injection. What is the final percent strength (w/v)
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To calculate the molality of potassium iodide (KI), we need to do the following steps:
Step 1: Determine the moles of KI present in the solution.
Step 2: Calculate the mass of water.
Step 3: Calculate the molality.
Let's break down each step:
Step 1: Determine the moles of KI present in the solution.
Molar mass of KI = 39.0983 g/mol (potassium) + 126.90 g/mol (iodine) = 166.9983 g/mol
To find the moles of KI, divide the given mass by the molar mass:
moles of KI = mass of KI / molar mass of KI
moles of KI = 0.520 g / 166.9983 g/mol
Step 2: Calculate the mass of water.
Given density of water = 1.000 g/mL
Volume of water = 120.0 mL
Mass of water = density × volume
Mass of water = 1.000 g/mL × 120.0 mL
Step 3: Calculate the molality.
Molality (m) is defined as the moles of solute divided by the mass of the solvent in kilograms.
Convert the mass of water from grams to kilograms:
mass of water (kg) = mass of water (g) / 1000
Now, we can calculate the molality using the formula:
molality (m) = moles of KI / mass of water (kg)
Plugging in the values we've found:
molality (m) = (0.520 g / 166.9983 g/mol) / (mass of water (g) / 1000)
molality (m) = (0.520 g / 166.9983 g/mol) / ((1.000 g/mL × 120.0 mL) / 1000)
After performing the calculations, the resulting value will give you the molality of the potassium iodide solution.