Question #2fed5

1 Answer
Jun 18, 2017

Answer:

The mixture contains 11.8 % #"NaCl"# by mass.

Explanation:

Let #x = "mass of NaCl"#

Then #(10.2 - x) = "mass of sucrose"#.

The formula for osmotic pressure #Pi# is

#color(blue)(bar(ul(|color(white)(a/a)Pi = icRTcolor(white)(a/a)|)))" "#

where

#i =# the van't Hoff #i# factor
#c =# the molar concentration of a component
#R =# the universal gas constant
#T =# the kelvin temperature

We have two solutes, #"NaCl"# and sucrose.

For #"NaCl"#:

#i = 2#

#"Moles of NaCl" = x/58.44color(white)(l) "mol" = "0.017 11"x color(white)(l)"mol"#

#c = ("0.017 11"color(white)(l) "mol")/"0.250 L" = "0.068 45"x color(white)(l)"mol/L"#

#T = "(23 + 273.15) K = 296.15 K"#

#Pi_text(NaCl) = 2 × "0.068 45"x color(red)(cancel(color(black)("mol·L"^"-1"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 296.15 color(red)(cancel(color(black)("K"))) = 3.327xcolor(white)(l) "atm"#

For sucrose:

#i = 1#

#"Moles of sucrose" = (10.2-x)/342.30 "mol" = ("0.029 80 - 0.002 921"x) color(white)(l)"mol"#

#c = (("0.029 80 - 0.002 921"x) color(white)(l)"mol")/"0.250 L" = ("0.1192 - 0.011 69"x) color(white)(l)"mol/L"#

#Pi_text(sucrose) = 1 × ("0.1192 - 0.011 69"x) color(red)(cancel(color(black)("mol·L"^"-1"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 296.15 color(red)(cancel(color(black)("K"))) = 24.30("0.1192 - 0.011 69"x) color(white)(l)"atm" = (2.897 - 0.2841x) color(white)(l)"atm"#

#Pi_text(NaCl) + Pi_text(sucrose) = Pi_text(total)#

#3.327x color(red)(cancel(color(black)("atm"))) + (2.897 - 0.2841x) color(red)(cancel(color(black)("atm"))) = 6.55 color(red)(cancel(color(black)("atm")))#

#3.327x + 2.897 - 0.2841x = 6.55#

#3.043x = 3.653#

#x = 3.653/3.043 = 1.20#

Mass of #"NaCl"# = 1.20 g.

#"Mass % NaCl" = (1.20 color(red)(cancel(color(black)("g"))))/(10.2 color(red)(cancel(color(black)("g")))) × 100 % = 11.8 %#