Question

For Questions 1 through 4 - Consider a solution prepared by combining 45.0 g of BaCl2 with enough water to prepare a 750.0 mL of solution. Assume that the volume of water added was 748 mL. Density of water is 1.00 g/mL. Answer questions 1 through 4. 1) What is the molarity of BaCl2 in the solution?

Answer 1

Answer: The molality of barium chloride solution is 0.289 m

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 748 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{748mL}\\\\\text{Mass of water}=(1g/mL\times 748mL)=748g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (barium chloride) = 45.0 g

$M_{solute}$ = Molar mass of solute (barium chloride) = 208.23  g/mol

$W_{solvent}$ = Mass of solvent (water) = 748 g

Putting values in above equation, we get:

$\text{Molality of barium chloride solution}=\frac{45\times 1000}{208.23\times 748}\\\\\text{Molality of barium chloride solution}=0.289m$

Hence, the molality of barium chloride solution is 0.289 m