Question

Bam added 8.91 g of CsCl to 34.29 mL of H2O. Calculate the concentration of Bam's solution in units of g solute/100 g solvent.

Answer 1

The concentration of the solution is  $26.1 g CsCl/100 g H_2O$.

We have  34.29 mL of water, whose density is 0.9975 g/mL. The corresponding mass of water is:

$34.29 mL \times \frac{0.9975g}{mL} = 34.20 g$

8.91 g of CsCl (solute) are in 34.20 g of water (solvent). We can calculate the concentration of Bam's solution in units of g solute/100 g solvent, using the following expression.

$C = \frac{mass\ solute }{mass\ solvent } \times 100 = \frac{8.91 g }{34.20 g } \times 100 = 26.1 g CsCl/100 g H_2O$

The concentration of the solution is  $26.1 g CsCl/100 g H_2O$.

You can learn more about solutions here: brainly.com/question/3055787