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58.5g of sodium iodide are dissolved in 0.5dm³. what concentration would this be?

User RussNS
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Final answer:

To find the concentration of the sodium iodide solution, divide the mass of sodium iodide (58.5g) by its molar mass (149.89 g/mol) to determine the moles, then divide by volume (0.5 liters) to find the molarity, which is 0.78 M.

Step-by-step explanation:

Calculating the Concentration of a Sodium Iodide Solution

To calculate the concentration of sodium iodide dissolved in water, we first need to know the molar mass of sodium iodide (NaI). The molar mass of NaI is approximately 149.89 g/mol. Using this information, we can determine the number of moles of NaI in 58.5g by dividing the mass by the molar mass.

Number of moles = mass (g) / molar mass (g/mol)

Number of moles = 58.5 g / 149.89 g/mol

Number of moles = 0.390 mol (rounded to three significant figures)

Next, we convert the volume from dm³ to liters, as 1 dm³ is equal to 1 liter. Therefore, the volume of the solution is 0.5 liters. Now we can calculate the concentration in moles per liter (mol/L), also known as molarity (M).

Concentration (M) = number of moles / volume (L)

Concentration (M) = 0.390 mol / 0.5 L

Concentration (M) = 0.78 M

Thus, the concentration of the sodium iodide solution is 0.78 mol/dm³.

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