Question

A mixture of NaCl and NaBr has a mass of 2.00 g and is found to contain 0.78 g of Na.

What is the mass of NaBr in the mixture? Express your answer using two significant figures.

Answer 1

Given that:

Mass of Mixture = 2.00 g

Mass of Na = 0.78 g

Let x be the moles of NaCl and y be the moles of NaBr

Molar mass of NaCl = 58.453 g/mol

Molar mass of NaBr = 102.9 g/mol

Thus, Mass of NaCl in mixture = moles * molar mass = x mol * 58.453g/mol = 58.453. x g

Mass of NaBr in mixture = moles * molar mass = y mol * 102.9g/mol = 102.9.y g

Now, total mass of mixture = mass of NaCl + mass of NaBr

2.0g = 58.453x g + 102.9y g . -----------------(A)

1 mol of each NaCl and NaBr contains 1 mol of Na respectively.

Molar mass of Na = 23 g/mol

THus, total mass of Na = (Moles of NaCl*molar mass of Na) + (moles of NaBr* molar mass of Na)

0.78 g = (x mol * 23g/mol) + ( y mol* 23g/mol) = 23(x+y) g

x + y = 0.0339

x = 0.0339-y

substituting this value of x in A, we get

2 g = 58.453 (0.0339-y) g + 102.9y g = 1.982 g - 58.453y g + 102.9y g

0.018 = 44.447y g

y = 0.000405 mol

Thus, x = 0.0339-y = 0.0339- 0.000405 mol = 0.0305 mol

Moles of NaBr in mixture =  0.000405 mol

Mass of NaBr =  0.000405 mol* 102.9 g/mol = 0.042g