Please provide detailed solutions and/or links. 1.) A 250-mL of standard solutio

Question

Please provide detailed solutions and/or links.

1.) A 250-mL of standard solution of 500.0 ppm Cu2+ (atomic mass: 63.55 g/mol) is available to prepare a 50-mL solution of 100.0 ppm. ______mL of the standard solution is needed to transfer to a ______mL volumetric flask before adding _________to the etched mark. The resultant solution has a molar concentration [Cu2+] of ___________M.

2.) 0.5000 g of dried Na2SO4 (F.W. 142.04 g/mol) is weighed out and transferred to a 250-mL volumetric flask, dissolving and diluting to the etched mark with deionized water (DI water). 10.00 mL of the above solution is then transferred with a _____________to a 100-mL volumetric flask to form a solution of ___________M Na+ after diluted with DI water to the mark. The same solution also contain [SO4 2- ] of __________M.

3.) A 0.500 g sample of a zinc ore was dissolved and the Zn (At Wt = 65.37) was precipitated as the phosphate, filtered, and weighed as Zn2P2O7 (MW = 304.8). If the Zn2P2O7 weighed 0.1090 g, _______% Zn can be calculated in the sample.

Explanation / Answer

Ans. #1. Using           C1V1 (original std. solution) = C2V2 (diluted solution)

Or, 500.0 ppm x V1 = 100.0 ppm x 50.0 mL

Or, V1 = (100.0 ppm x 50.0 mL) / 500.0 ppm = 10.0 mL

# [Cu2+] in resultant solution = 100 ppm = 100 mg/ L = 0.100 g / L

                                                = (0.100 g / 63.55 g mol-1) / L

                                                = 0.00157 mol/ L

                                                = 0.00157 M

A 250-mL of standard solution of 500.0 ppm Cu2+ (atomic mass: 63.55 g/mol) is available to prepare a 50-mL solution of 100.0 ppm. 10.0 mL of the standard solution is needed to transfer to a 50.0 mL volumetric flask before adding deionized water to the etched mark. The resultant solution has a molar concentration [Cu2+] of 0.00157 M.

#2. # Step 1: Preparation of standard solution:

Moles of Na2SO4 = 0.5000 g /(142.04 g/ mol) = 0.0035201 mol

Now,

            [Na2SO4] = moles / Vol. in liters = 0.0035201 mol / 0.250 L = 0.0140804 M

# Step 2: Using        C1V1 (original std. solution) = C2V2 (diluted solution)

C2 = (0.0140804 M x 10.0 mL) / 100.0 mL = 0.00140804 M

# Since 1 mol Na2SO4 dissociates to yield 1 mol SO42-, and 2 mol Na+           -

            [SO42-] = [Na2SO4] = 0.00140804 M

            [Na+] = 2 x [Na2SO4] = 0.00281608 M

# 0.5000 g of dried Na2SO4 (F.W. 142.04 g/mol) is weighed out and transferred to a 250-mL volumetric flask, dissolving and diluting to the etched mark with deionized water (DI water). 10.00 mL of the above solution is then transferred with a volumetric pipette to a 100-mL volumetric flask to form a solution of 0.00281608 M Na+ after diluted with DI water to the mark. The same solution also contain [SO4 2- ] of 0.00140804 M.

#3: Moles of Zn2P2O7 = 0.1090 g / (304.8 g/ mol) = 3.576115 x 10-4 mol

# 1 mol Zn2P2O7 consists of 2 mol Zn.

So, moles of Zn in sample = 2 x moles of Zn2P2O7 = 2 x 3.576115 x 10-4 mol

                                                = 7.15223 x 10-4 mol

# Now, mass of Zn in sample = 7.15223 x 10-4 mol x (65.37 g/ mol) = 0.0468 g

# % Zn in sample = (Mass of Zn / Mass of sample) x 100

                                    = (0.0468 g / 0.500 g) x 100

                                    = 9.36 %

# A 0.500 g sample of a zinc ore was dissolved and the Zn (At Wt = 65.37) was precipitated as the phosphate, filtered, and weighed as Zn2P2O7 (MW = 304.8). If the Zn2P2O7 weighed 0.1090 g, 9.36 % Zn can be calculated in the sample.

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