Question #31c13
1 Answer
Here's what I got.
Explanation:
The information you provided here is a bit sketchy, but it does allow you to find an expression for the molality of the solution.
Molality is defined as the number of moles of solute present for every
Now, you know that you're at room temperature, so assume that the temperature is equal to
You'll find that
#rho_"water" = "0.9982 g mL"^(-1)#
https://www.simetric.co.uk/si_water.htm
This means that the mass of the solvent will be equal to
#100 color(red)(cancel(color(black)("mL"))) * "0.9982 g"/(1color(red)(cancel(color(black)("mL")))) = "99.82 g"#
To calculate the molality of the solution, you must use the molar mass of the solute to convert its mass to moles, then determine the number of moles of solute present for every
If you take
#40 color(red)(cancel(color(black)("g"))) * "1 mole"/(M_Mcolor(red)(cancel(color(black)("g")))) = (40/M_M)# #"moles"#
The number of moles of solute present for every
#10^3 color(red)(cancel(color(black)("g water"))) * ((40/M_M)color(white)(.)"moles")/(99.82color(red)(cancel(color(black)("g water")))) = (400.72/M_M)# #"moles"#
Therefore, the solution has a molality of
#"molality" = (400.72/M_M)# #"mol kg"^(-1)#
Keep in mind that you must round the actual value to one significant figure, the number of sig figs you have for your values.