1) A student weighs out a 5.75 g sample of CuBr2, transfers it to a 125 mL volum

Question

1) A student weighs out a 5.75 g sample of CuBr2, transfers it to a 125 mL volumetric flask, adds enough water to dissolve it and then adds water to the 125 mL tic mark. What is the molarity of copper(II) bromide in the resulting solution? M

2)

How many mL of a 0.231 M aqueous solution of sodium bromide, NaBr, must be taken to obtain 3.19 grams of the salt?

mL

3)

An aqueous solution of copper(II) bromide, CuBr2, is made by dissolving 21.8 grams of copper(II) bromide in sufficient water in a300 mL volumetric flask, and then adding enough water to fill the flask to the mark.

What is the weight/volume percentage of copper(II) bromide in the solution?  %

4)

How many grams of sodium bromide, NaBr, are required to make a 8.81 % w/v aqueous solution in a 200 mL volumetric flask?    grams

Explanation / Answer

1) molarity of CuBr2 solution = 5.75 g/223.37 g/mol x 0.125 L = 0.206 M

2) 0.231 M NaBr solution has = 0.231 mol/L x 102.894 g.mol = 23.77 g NaBr in 1 L of solution

So to obtain 3.19 g NaBr we would require = 3.19 g x 1000 ml/23.77 g = 134.203 ml of solution

3) wt of CuBr2 = 21.8 g

Volume of solvent = 300 ml

So, wt/Vol % of CuBr2 in solution = (21.8/300) x 100 = 7.26%

4) To prepare a 8.81% (w/v) NaBr solution,

grams of NaBr required = 8.81 x 200/100 = 17.62 g

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