Final answer:
To find the molarity of 21.3 g of MgS in 765 mL of solution, we calculate the moles of MgS and divide by the volume in liters. The molarity is approximately 0.494 M which, when rounded to one significant figure, matches answer choice (c) 0.5 M.
Step-by-step explanation:
To calculate the molarity of 21.3 g of MgS in 765 mL of solution, we first need to determine the molar mass of MgS. The molar mass of magnesium (Mg) is approximately 24.305 g/mol, and the molar mass of sulfur (S) is approximately 32.065 g/mol. Therefore, the molar mass of MgS is 24.305 g/mol + 32.065 g/mol = 56.370 g/mol.
Next, we calculate the number of moles of MgS as follows:
Moles of MgS = mass (g) / molar mass (g/mol)
Moles of MgS = 21.3 g / 56.370 g/mol ≈ 0.378 moles
To find the molarity, we need to convert the volume of the solution from milliliters to liters:
Volume in liters = 765 mL * (1 L/1000 mL) = 0.765 L
Molarity (M) = moles of solute / liters of solution
Molarity = 0.378 moles / 0.765 L ≈ 0.494 M
When rounding to one significant figure, the closest answer is (c) 0.5 M.