All problems assume P = 1.0 atm and T = 25 0 C, unless otherwise noted 1. To fiv
ID: 800422 • Letter: A
Question
All problems assume P = 1.0 atm and T = 25 0C, unless otherwise noted
1. To five significant figures, how much sodium, Na+, in grams, is in 1.50 L of a 0.023-molal solution of NaNO3 and how much sodium, in grams is in 1.50 L of a 0.0230-molar (M) solution of NaNO3? Using the equation below, what is the difference (%) between the molality and molarity of the solution? (12)<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
difference (%) = (value 1 - value 2)/(value 1 +value 2)(100)
2. Repeat problem 1 at 10 0C (8)
Explanation / Answer
Given 1.50 L of 0.0230 M solution
moles = molarity x volume
moles =0.0230 x 1.50
moles = 0.0345
mass = moles x M.w
mass of Na+ = 0.0345 x 23
mass of Na+ = 0.7935 grams
Value 2 = 0.7935 grams
PM = dRT
d= PM /RT
d= 101325 x 18 / 8.314 x 298
d= 736.144 g/L
mass of water = 0.736144 x 1.5 = 1.104
moles = molality x mass of solvent
moles = 0.023 x 1.104
moles of Na+ = 0.02539
mass of Na + = 23 x 0.02539
mass of Na+ = 0.5841
value 1 =0.5841 gram
% diff = ( 0.5841 - 0.7935 ) / (0.5841 + 0.7935 ) x 100
% diff = 15.20
2) PM = dRT
d= PM /RT
d= 101325 x 18 / 8.314 x 283
d= 775.162 g/L
mass of water = 0.7775162 x 1.5 = 1.162
moles = molality x mass of solvent
moles = 0.023 x 1.162
moles of Na+ = 0.026743
mass of Na + = 23 x 0.026743
mass of Na+ = 0.6151
value 1 =0.6151 gram
% diff = ( 0.6151 - 0.7935 ) / (0.6151 + 0.7935 ) x 100
% diff = 12.66