Question

What is the concentration (in m) of CaBr2 in a solution prepared by dissolving 5.73 g of CaBr2 in 457 g of water? What is the mole fraction of CaBr2 in this solution?
molality: ?
mole fraction:?
Answer: ______________________
Answer: ______________________

Answer 1

1)

$$c_m(CaBr_2) = {n(CaBr_2) \over m(H_2O)} = {m(CaBr_2) \over M(CaBr_2)m(H_2O)}$$

$$c_m(CaBr_2) = {5.73g \over 199.89g/mol *0.457kg} = 0.063 mol/kg$$

2)

$$\chi(CaBr_2) = {n(CaBr_2)\over n(CaBr_2)+n(H_2O)}$$

$$\chi(CaBr_2) = {5.73g \over 199.89g/mol(5.73g/199.89g/mol + 457g/18.015g/mol)}$$

$$\chi(CaBr_2) = 0.0011$$