Question

ltihium iodide has a solubility of 167.5/100 g h2o . an unsaturated stock solution of lithium iodide is prepared by dissolving 180.00 g of lithium iodide in 250 ml of water the solution has a vinal volume of 385.0 ml (d_water=1.000 g/ml , d_solution - 1.117g/ml

what additional mass of lithium iodide must be added to the original stock soloution to create saturated lithium iodide solution?

Answer 1

Volume of water taken = 250 mL; density of water = 1.000 g/mL.

Therefore, mass of water taken = (250 mL)*(1.000 g/mL) = 250.00 g.

Final volume of solution = 385 mL; density of solution = 1.117 g/mL.

Therefore, final mass of solution = (385 mL)*(1.117 g/mL) = 430.045 g.

Mass of lithium iodide in the final solution = (mass of solution) – (mass of water taken) = (430.045 g) – (250.000 g) = 180.045 g.

100 g water contains 167.5 g lithium iodide.

Therefore, 250 g water must contain (167.5 g)*(250 g/100 g) = 418.75 g lithium iodide.

Mass of lithium iodide to be added = (418.75 g) – (180.045 g) = 238.705 g (ans).