Final answer:
To prepare a 250 mL solution of 35% CaCl₂, assuming a solution density similar to water, 87.5 grams of CaCl₂ are needed.
Step-by-step explanation:
To determine how many grams of CaCl₂ are needed to make a 250 mL solution of 35% CaCl₂, we first need to calculate the amount of CaCl₂ in the solution.
35% solution means that 35 g of CaCl₂ is present in 100 mL of solution. So, in 250 mL of solution, there would be:
35 g/100 mL x 250 mL = 87.5 g of CaCl₂
To calculate how many grams of CaCl₂ are needed to make a 250 mL solution of 35% CaCl₂, we use the formula:
Mass of Solute (g) = Volume of Solution (L) × Density of Solution (g/mL) × Percent Concentration
Assuming the density of the solution is close to that of water (since a density value is not provided), which is 1 g/mL, the calculation is as follows:
Mass of Solute (g) = 250 mL × 1 g/mL × 0.35 = 87.5 grams of CaCl₂.
Therefore, you would need 87.5 grams of CaCl₂ to prepare a 250 mL solution of 35% CaCl₂.