To answer the above question, you must calculate the osmotic pressures of all the solutions.
The formula for osmotic pressure is
#color(blue)(|bar(ul(color(white)(a/a) Π = MRT color(white)(a/a)|)))" "#
In each case, let's assume that we have 1 L of solution at 25 °C.
6 % urea solution
Assume that density = 1.0 g/mL.
Then mass of solution = 1000 g.
#"Mass of urea" = 1000 color(red)(cancel(color(black)("g solution"))) × "6 g urea"/(100 color(red)(cancel(color(black)("g solution")))) = "60 g urea"#
#"Moles of urea" = 60 color(red)(cancel(color(black)("g urea"))) × "1 mol urea"/(60.06 color(red)(cancel(color(black)("g urea")))) = "1.00 mol"#
#"Molarity" = "moles"/"litres" = "1.00 mol"/"1 L" = "1 mol/L"#
#Π = MRT = 1 .00 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"))) × 298 color(red)(cancel(color(black)("K"))) = "24 atm"#
#(1)# 6 % glucose
Mass of glucose = 60 g
#"Moles of glucose" = 60 color(red)(cancel(color(black)("g glucose"))) × "1 mol glucose"/(180 color(red)(cancel(color(black)("g glucose")))) = "0.33 mol glucose"#
#"Molarity" = "moles"/"litres" = "0.33 mol"/"1 L" = "0.33 mol/L"#
#Π = MRT = 0.33 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"))) × 298 color(red)(cancel(color(black)("K"))) = "8.1 atm"#
#(2)# 25 % glucose
Assume that density = 1.1 g/mL
Then mass of solution = 1100 g.
#"Mass of glucose" = 1100 color(red)(cancel(color(black)("g solution"))) × "25 g glucose"/(100 color(red)(cancel(color(black)("g solution")))) = "275 g glucose"#
#"Moles of glucose" = 275 color(red)(cancel(color(black)("g glucose"))) × "1 mol glucose"/(180 color(red)(cancel(color(black)("g glucose")))) = "1.53 mol glucose"#
#"Molarity" = "moles"/"litres" = "1.53 mol"/"1 L" = "1.53 mol/L"#
#Π = MRT = 1 .53 "mol·L"^"-1" × "0.082 06" "L"·"atm"·"K"^"-1""mol"^"-1" × 298 "K" = "37 atm"#
#(3)# 1 mol/L glucose
#Π = MRT = 1 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"))) × 298 color(red)(cancel(color(black)("K"))) = "24 atm"#
#(4)# 0.05 mol/L glucose
#Π = MRT = 0.05 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"))) × 298 color(red)(cancel(color(black)("K"))) = "1.2 atm"#
6 % urea has the same osmotic pressure as #(3)# 1 mol/L glucose.
The two solutions are isotonic.
