Question

How many moles of AgCl are contained in 244 ml of 0.135 m AgCl solution? The density of the solution is 1.22 g/mL. O 3.49 x 10-2 mol AgCI 4.31 x 10-2 mol AgCI 2.32 x 10-mol AgCI O 1.85 x 10 2 mol AgCI O 3.29 x 10 2 mol AgCI Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining

Answer 1

Solution

Molality of AgCl=0.135 m

Density of 0.135 m solution = 1.22 g/mL

Volume of solution in mL = 244 mL

mass of solution in g = volume of solution x Density = 244 mL x 1.22 g/mL

mass of solution in g = 297.68 g (i.e. 297.68 / 1000 = 0.298 kg)

we know

molality = number of mole solute / volume of solution in kg

0.135 mol /kg = number of mole AgCl / 0.298 kg

Number of mole AgCl = 0.135 mol /kg x 0.298 kg

Number of mole AgCl = 0.04023 moles

Number of mole AgCl = 4.023 x 10^-2 moles

so ans 2 is correct

Answer 2

The correct answer is 3.49e-2