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Chemical Concentrations and Calculations
Question:
Please give a proper explanation and provide answers for all:
Q1:
You weigh out 105 mg of Winterium flakide (Formula WiFl2, mw = 90.380 g/mol) and dissolve it in 2.3 L of water. What is the concentration of Winterium (mw 24.500 g/mol) to three decimal places in ppm?
Q2:
You weigh out 70 mg of Icium Chrystalide (Formula Ic2Ch3, mw = 93.494 g/mol) and dissolve it in 1.7 L of water. What is the concentration of iodine (mw 24.500 g/mol) to three decimal places in ppm.
Q3:
You are asked to make 1.9 L of a 110 ppm solution of Icium (mw 23.375 g/mol) using Icium Chrystalide (Formula Ic2Ch3, mw = 84.152 g/mol). How much Winterium Flakide do you need to weigh out to achieve this? Report your answer to 3 decimal places.
Q4:
You analyze standard solutions of sodium, create a calibration curve and obtain an equation of the line: y = 279.872x + -2.978. You take an unknown and dilute it: 14 mL into a 796.1 mL volumetric flask and filled to the mark. You analyze it and find an emission intensity of 65. What is the concentration of the original unknown in ppm to 3 decimal places.
Answer:
1. Concentration of Winterium in ppm after dissolving in water:
a. Calculate moles of Winterium Flakide (WiFl2):
moles of WiFl2=(0.105 g)/(90.380 g/mol)
≈0.001162 mol
b. Calculate concentration in mol/L:
concentration of WiFl2=(0.001162 mol)/(2.3 L)
≈0.000505 mol/L
c. Convert concentration to ppm:
concentration of WiFl2 in ppm=(0.000505 mol/L)×10^6
≈505 ppm
2. Concentration of Icium in ppm after dissolving in water:
a. Calculate moles of Icium Chrystalide (Ic2Ch3):
moles of Ic2Ch3=(0.07 g)/(93.494 g/mol)
≈0.000748 mol
b. Calculate concentration in mol/L:
concentration of Ic2Ch3=(0.000748 mol)/(1.7 L)
≈0.000440 mol/L
c. Convert concentration to ppm:
concentration of Ic2Ch3 in ppm=(0.000440 mol/L)×10^6
≈440 ppm
3. Amount of Winterium Flakide needed to make a 110 ppm Icium solution:
a. Calculate moles of Icium needed:
moles of Icium=110×1.9/10^6
≈0.000209 mol
b. Calculate the weight of Winterium Flakide needed:
weight of WiFl2=0.000209 mol×90.380 g/mol
≈0.0189 g
3 decimal places:
≈0.019 g
≈0.019g
4. Concentration of the original unknown sodium solution:
a. Substitute the given values into the equation of the line: y=279.872x−2.978
concentration of sodium in unknown=(65+2.978)/279.872
≈0.233 mol/L
b. Calculate the dilution factor:
dilution factor=796.1/14
≈56.864
c. Calculate the concentration of the original unknown in ppm: concentration of original unknown in ppm=0.233×56.864×10^6≈13248 ppm
