Question

Calcium chloride, CaCl2, is commonly used as an electrolyte in sports drinks and other beverages, including bottled water. A solution is made by adding 6.50 g of CaCl2 to 60.0 mL of water at 25∘C. The density of the solvent at that temperature is 0.997 g/mL. Calculate the mole percent of CaCl2 in the solution.

Answer 1

Answer:

Mole percent of $CaCl_{2}$ in solution is 1.71%

Explanation:

Number of moles of a compound is the ratio of mass to molar mass of the compound.

Molar mass of $CaCl_{2}$ = 110.98 g/mol

Molar mass of $H_{2}O$ = 18.02 g/mol

Density is the ratio of mass to volume

So, mass of 60.0 mL of water = $(60\times 0.997)g=60.8g$

Hence, 6.50 g of $CaCl_{2}$ = $\frac{6.50}{110.98}moles$ of $CaCl_{2}$ = 0.0586 moles of $CaCl_{2}$

60.8 g of $H_{2}O$= $\frac{60.8}{18.02}moles$ of $H_{2}O$ = 3.37 moles of $H_{2}O$

So, mole percent of $CaCl_{2}$ in solution = \frac{n_{CaCl_{2}}}{n_{total}}\times 100% = $\frac{0.0586}{0.0586+3.37}\times 100$% = 1.71%