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.

Answer 1

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
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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.

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