What mass of #AgNO_3# would be required to prepare a 0.250 molal solution in 125 g of water?

1 Answer
Jul 9, 2016


#"5.31 g AgNO"_3#


Your strategy here will be to

  • use the molality and mass of water to calculate how many moles of solute you have in your solution
  • use the molar mass of silver nitrate to convert the moles to grams

As you know, molality is defined as moles of solute, which in your case is silver nitrate, per kilogram of solvent. This means that a solution's molality essentially tells you the number of moles of solute present in #"1 kg"# of solvent.

In your case, a #"0.250 molal"# solution would contain #0.250# moles of solute for every #"1 kg"# of solvent.

The problem tells you that you have #"125 g"# of solvent available. Use the solution's molality as a conversion factor to calculate how many moles of silver nitrate it must contain -- do not forget to convert the mass of solvent from grams to kilograms

#125 color(red)(cancel(color(black)("g"))) * (1 color(red)(cancel(color(black)("kg"))))/(10^3color(red)(cancel(color(black)("g")))) * overbrace("0.250 moles AgNO"_3/(1color(red)(cancel(color(black)("kg solvent")))))^(color(blue)("= 0.250 molal")) = "0.03125 moles AgNO"_3#

Now, you know that silver nitrate has a molar mass of #"169.87 g mol"^(-1)#, which means that one mole of silver nitrate has a mass of #"169.87 g"#.

You can thus say that #0.03125# moles of silver nitrate will have a mass of

#0.03125 color(red)(cancel(color(black)("moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("5.31 g")color(white)(a/a)|)))#

The answer is rounded to three sig figs.