What is the % by mass of Mg2+(aq) in the 0.00500 M MgCl2(aq) solution from Quest

Question

What is the % by mass of Mg2+(aq) in the 0.00500 M MgCl2(aq) solution from Question 1(Molarity of Mg+2 = 0.00500 M). Assume the solution density is 1.0 g/mL, like water. (3 significant digits, unit %) 1 points

QUESTION 6 What is the molality of the Mg2+(aq) in the 0.00500 M MgCl2(aq) solution from Question 1(Molarity of Mg+2 = 0.00500 M). Assume the solution density is 1.0 g/mL, like water. (3 significant digits, unit m) 1 points

QUESTION 7 Stu and Noke are working with a red food coloring, Red #40. The stock solution has a concentration of 0.0880 M dye. Noke want to makes a concentration of 0.0088 M, exactly one tenth of the original red dye solution strength. What volume of the original 0.0880 M solution should she measure into a 100.0 volumetric flask, to be diluted with water up to the line of the flask? (3 significant digits, unit mL) 1 points

QUESTION 8 Stu dilutes the stock 0.0880 M red dye solution (same as the previous question) by measuring 25.00 mL with a pipet into a 100.0 mL volumetric flask and then adding water up to the line on the flask. Calculate the new concentration of his diluted solution. (Round to 3 significant digits, unit M) 1 points

QUESTION 9 The next three questions focus on a 26 ppm Ag+(aq) solution that Kemmi and Doc are using. First, calculate the molarity of the silver cation (Ag+) in the solution. Assume the solution density is the same as pure water. (2 significant digits, unit M) 1 points

QUESTION 10 Kemmi dilutes the 26 ppm Ag+(aq) solution by twice pipetting 15 mL of solution into a 100 mL volumetric flask. She fills the remaining volume with water. Use M1V1 = M2V2 with M1 equal to 26 ppm. Solve for the diluted concentration (M2) in units of ppm. (2 significant digits, unit ppm) 1 points

QUESTION 11 Doc wants to make a 10.4 ppm Ag+(aq) solution from the original 26 ppm solution. He has a 25 mL volumetric flask available. Use M1V1 = M2V2 with the M1 and M2 values in units of ppm, to predict which volumetric pipet Doc should use to make his dilution. (2 significant digits, unit mL)

Explanation / Answer

1) Concentration of MgCl2 = 0.00500 M; density of solution = 1.00 g/mL.

Therefore, 1 L = 1000 mL solution contains 0.00500 mole MgCl2.

Molar mass of Mg = 24 g/mole; therefore, mass of Mg2+ = (0.00500 mole)*(24 g/mol) = 0.12 g.

Mass of the solution = (1000 mL)*(1 g/mL) = 1000 g.

Therefore, 1000 g solution contains 0.12 g Mg2+.

Percent Mg in the solution = (0.12 g)/(1000 g)*100 = 0.012% (ans).

6) Molarity of MgCl2 = 0.00500 mole; mass of the solution = 1000 g.

Molar mass of MgCl2 = (24 + 2*35.5) g/mol = 95 g/mol.

Therefore, mass of MgCl2 in the solution = (0.00500 mole)*(95 g/mol) = 0.475 g.

The mass of solvent (water) = (mass of solution) – (mass of solute) = (1000 g) – (0 in the solution = (0.00500 mole)*(95 g/mol) = 0.475 g.

The mass of solvent (water) = (mass of solution) – (mass of solute) = (1000 g) – (0.475 g) = 999.525 g = (999.525 g)*(1 kg/1000 g) = 0.999525 kg.

Moles of Mg2+ (deduced above) = 0.00500 mole

Therefore, molality of Mg2+ = moles of Mg2+/kg of solvent = (0.00500)/(0.999525) m = 0.005002 m 0.005 m (ans).

7) Use the dilution equation: M1*V1 = M2*V2

M1 = molar concentration of the more concentrated (stock solution); M2 = molar concentration of the less concentrated (dilute) solution; V1 = volume of the stock solution required/taken and V2 = final volume of the dilute solution.

Plug in values. Write

(0.0880 M)*V = (0.00880 M)*(100 mL)

===> V = (0.00880*100)/(0.0880) mL = 10 mL

Therefore, 10 mL of the stock solution will have to be diluted to 100 mL to get the desired concentration.

8) Here, it is given V1 = 25 mL and V2 = 100 mL. We are supposed to find out M2. Plug in values:

(0.0880 M)*(25 mL) = M*(100 mL)

===> M = (0.0880*25)/(100) M = 0.022 M

The new concentration is 0.022 M (ans).

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