Question

Answer each of the following questions in your own words; show all work, and express numeric values to the appropriate level of significance. 1. Calculate the molarity of a solution comprised of 207.4 g silver nitrate in 1.26 x 101 mL cf solution. If 100.0 mL of this solution are diluted to a volume of 550.0 mL, what is the new concentration? 2. If 360.4 mL of a 0.0290 M solution of sodium sulfate are diluted to a total volume of 500. mL, what is the new concentration? How many grams of sodium sulfate are present in this solution?

Answer 1

Ans 1

Moles of silver nitrate (AgNO3) = mass/molecular weight

= 207.4 g / 169.87 g/mol

= 1.22 mol

Volume = 1.26*10^4 mL x 1L /1000 mL

= 12.6 L

Molarity = moles /Volume

= 1.22 mol / 12.6 L

= 0.0968 M

After dilution

Total volume = 12.6 + (0.550 - 0.100) = 13.05 L

Molarity = 1.22 mol / 13.05 L

= 0.0934 M