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Explain how you would make 675 ml of a 0.400 m barium iodide solution.
Accepted Answer
Materials:
* Barium iodide (BaI2)
* Deionized water
* Graduated cylinder
* Volumetric flask (675 mL)
Steps:
1. Calculate the mass of BaI2 needed:
Moles of BaI2 = Concentration (M) x Volume (L)
Moles of BaI2 = 0.400 M x 0.675 L
Moles of BaI2 = 0.270 moles
Mass of BaI2 = Moles of BaI2 x Molar mass of BaI2
Mass of BaI2 = 0.270 moles x 331.85 g/mol
Mass of BaI2 = 89.80 g
2. Weigh out the calculated mass of BaI2.
3. Transfer the BaI2 to a 675 mL volumetric flask.
4. Add about 500 mL of deionized water to the flask and swirl to dissolve the BaI2.
5. Fill the flask to the 675 mL mark with deionized water.
6. Stopper the flask and invert several times to mix the solution thoroughly.
The resulting solution will be 675 mL of 0.400 M barium iodide.
* Barium iodide (BaI2)
* Deionized water
* Graduated cylinder
* Volumetric flask (675 mL)
Steps:
1. Calculate the mass of BaI2 needed:
Moles of BaI2 = Concentration (M) x Volume (L)
Moles of BaI2 = 0.400 M x 0.675 L
Moles of BaI2 = 0.270 moles
Mass of BaI2 = Moles of BaI2 x Molar mass of BaI2
Mass of BaI2 = 0.270 moles x 331.85 g/mol
Mass of BaI2 = 89.80 g
2. Weigh out the calculated mass of BaI2.
3. Transfer the BaI2 to a 675 mL volumetric flask.
4. Add about 500 mL of deionized water to the flask and swirl to dissolve the BaI2.
5. Fill the flask to the 675 mL mark with deionized water.
6. Stopper the flask and invert several times to mix the solution thoroughly.
The resulting solution will be 675 mL of 0.400 M barium iodide.
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