Question
Calculate the mole fraction of a mixture of NaBr and KBr which is 0.981g of mass, given that we are treating the mixture with excessive AgNO3. 1.690g of AgBr is formed.
Answers
DrBob222
This is two unknowns so it requires two equations and solve them simultaneously. First, you must determine the grams NaBr and grams KBr.
Let X = grams NaBr
and Y = grams KBr
====================
equation 1 is X + Y = 0.981
equation 2 is obtained by converting grams X to grams AgBr and converting grams Y to grams AgBr. You know the total grams AgBr is 1.690. Equation 2 is
(mm stands for molar mass)
X(mmAgBr/mmNaBr)+(Y(mmAgBr/mmKBr) = 1.690.
Solve those two equations simultaneously for X and Y and that gives you grams NaBr and grams KBr.
mols NaBr = grams/molar mass = ?
mols KBr = grams/molar mass = ?
XNaBr = mols NaBr/total mols
XKBr = mols KBr/total mols.
Post your work if you stuck.
Let X = grams NaBr
and Y = grams KBr
====================
equation 1 is X + Y = 0.981
equation 2 is obtained by converting grams X to grams AgBr and converting grams Y to grams AgBr. You know the total grams AgBr is 1.690. Equation 2 is
(mm stands for molar mass)
X(mmAgBr/mmNaBr)+(Y(mmAgBr/mmKBr) = 1.690.
Solve those two equations simultaneously for X and Y and that gives you grams NaBr and grams KBr.
mols NaBr = grams/molar mass = ?
mols KBr = grams/molar mass = ?
XNaBr = mols NaBr/total mols
XKBr = mols KBr/total mols.
Post your work if you stuck.