Asked by John
What is the mass in grams of copper chloride dihydrate (CuCl2·2H2O) required to prepare 80.0 mL of a 0.10 M CuCl2 solution?
Answers
Answered by
Damon
Cu =63.5 * 1 = 63.5
Cl = 35.5 * 2 = 71
sum = 134.5
H = 1 * 4 = 4
O = 16 * 2 = 32
sum = 36
entire hydrated molecule therefore 170.5 g/mol
0.10 * 170.5 = 17.05 g/ L
17.05 g/L * 80 *10^-3 L = 13.6 grams of hydrated stuff
note mols CuCl2 same as mols CuCl2 2H2O, one Cu in each
Cl = 35.5 * 2 = 71
sum = 134.5
H = 1 * 4 = 4
O = 16 * 2 = 32
sum = 36
entire hydrated molecule therefore 170.5 g/mol
0.10 * 170.5 = 17.05 g/ L
17.05 g/L * 80 *10^-3 L = 13.6 grams of hydrated stuff
note mols CuCl2 same as mols CuCl2 2H2O, one Cu in each
Answered by
DrBob222
mols CuCl2 needed = M x L = 0.10 x 0.0800= ?
grams CuCl2 = mols CuCl2 x molar mass CuCl2.= ?
Then g CuCl2.2H2O = grams CuCl2 x (molar mass CuCl2.2H2O/molar mass CuCl2) = ?
Post your work if you get stuck.
grams CuCl2 = mols CuCl2 x molar mass CuCl2.= ?
Then g CuCl2.2H2O = grams CuCl2 x (molar mass CuCl2.2H2O/molar mass CuCl2) = ?
Post your work if you get stuck.
Answered by
DrBob222
@Damon--check your decimal point?
Answered by
Damon
Whoops, thanks, 1.36
Answered by
DrBob222
Easy to slip a decimal. :-)
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