Determine amount of NaBr dissolved in filtrate and solubility in methanol (g/L)

Question

Initial mass: 10.040 g

Final mass: 3.284

Vol of filtrate: 50 mL

Answer 1

The amount of NaBr dissolved in the filtrate would be equal to the initial mass of NaBr minus the final mass of NaBr. Assuming the final mass is in grams (g), the amount of NaBr in the filtrate is:

10.040 g - 3.284 g = 6.756 g

We can calculate the solubility of NaBr in methanol in g/L, now that we know 50 mL methanol was required to dissolve 6.756 g NaBr. All we have to do is convert the units to g/L. Since there are 1000 mL in 1 L, 50 mL equals 0.05 L (just divide by 1000 to go from mL to L). The solubility in g/L is (6.756 g)÷(0.05 L) = 135.12. If we want to round to the correct number of significant figures, the answer is 135.1 g/L.