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Question 10 Elemental sulfur can be used to control outbreaks of powdery mildew on grapes - VCE - SSCE Chemistry - Question 10 - 2009 - Paper 1

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Question 10 Elemental sulfur can be used to control outbreaks of powdery mildew on grapes. However, sulfur remaining on the grapes after harvest can be converted to... show full transcript

Worked Solution & Example Answer:Question 10 Elemental sulfur can be used to control outbreaks of powdery mildew on grapes - VCE - SSCE Chemistry - Question 10 - 2009 - Paper 1

Step 1

Determine the concentration of barium ions remaining in the 10.00 mL sample solution.

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Answer

To find the concentration of barium ions remaining in the sample solution, we first refer to the calibration curve using the absorbance of the solution.

Given that the absorbance measured is 0.11, we correlate this value with the linear relationship defined by the standard solutions:

  1. Locate the absorbance of 0.11 on the Y-axis of the calibration curve.
  2. Draw a horizontal line from this point until it intersects with the calibration curve.
  3. From the intersection, draw a vertical line down to the X-axis to read the corresponding concentration of barium ions.

Assuming the graph shows a concentration of approximately 21 mg/L: The concentration (CBaC_{Ba}) in the 10.00 mL sample solution is 21 mg/L.

Next, to find the mass of barium ions remaining:

Using the formula:

extMass=extConcentrationimesextVolume ext{Mass} = ext{Concentration} imes ext{Volume}

we convert the volume from mL to L:

extVolume=10.00mL=0.010L ext{Volume} = 10.00 mL = 0.010 L

Then calculate:

extMass=21extmg/Limes0.010extL=0.21extmg ext{Mass} = 21 ext{ mg/L} imes 0.010 ext{ L} = 0.21 ext{ mg}

Step 2

Determine the mass of barium ions, in mg, remaining in the 10.00 mL sample solution.

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Answer

Using the concentration calculated previously, the mass of the barium ions remains:

extMassBa=21extmg/Limes0.010extL=0.21extmg ext{Mass}_{Ba} = 21 ext{ mg/L} imes 0.010 ext{ L} = 0.21 ext{ mg}

Hence, the mass of barium ions remaining in the 10.00 mL sample solution is 0.21 mg.

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