Question

The solubility of solid W in water is: 1.72 g/100 mL at 0°C, 21.3/100 mL at 100°C. a) How many mL of boiling water are required to dissolve 154.0 g of W?(report to the nearest mL) If solution were cooled to 0°C, how many grams of W would crystallize out? (report to one decimal place) b) What is the percent recovery? (report to one decimal place) (Show calculations)

Answer 1

Answer:

a) Mass of solid W crystallized out 0°C is 141.564 g

b) 91.92% is the percent recovery.

Explanation:

The solubility of solid W in water at 100°C = 21.3 g/100 mL

Volume of water at 100C required to dissolve 154.0 g of W: x

$\frac{154.0 g}{x}=\frac{21.3 g}{100 mL}$

$x=\frac{154.0 g\times 100 mL}{21.3 g}=723.0 mL$

723.0 mL of water will dissolve 154.0 grams of solid W.

The solubility of solid W in water at 0°C = 1.72 g/100 mL

Mass of W soluble in 723.00 mL of water at  0°C = m

$\frac{m}{723.0 mL}=\frac{1.72 g}{100 mL}$

$m=\frac{1.72 g\times 723.0 mL}{100 mL}=12.436 g$

a) Mass of solid W crystallized out 0°C : 154.0 g- 12.436 g=141.564 g

b)  Percent recovery of solid at 0°C:

$\frac{141.564 g}{154.0 g}\times 100=91.92\%$

91.92% is the percent recovery.