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Grade 12Physical Chemistry

Molarity of 0.2m aqeous solution of nacl having density 5/4g/ml is

Profile image of Amit
7 Years agoGrade 12
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2 Answers

Profile image of Pooja
7 Years ago
Molality (m)= 0.2mdensity (d) = 5/4 g/mlWeight of solute = w gmWeight of solvent = W gmVolume of solution = V mlMolecular mass of NaCl = MM MM = 58.5 gm/molem = (w×1000)/W× MM0.2 × 58.5 = (w×1000)/W(w×1000) = 11.7W .....................(1)d = (w+W)/V5/4 = {(11.7 W/1000) + W}÷V5V/4 = 1011.7 W ÷10001250 V= 1011.7 W1250÷1011.7 = W/V .........................(2)M = (w×1000)/V×MMM = (m×W)/VM = 0.2 × 1250÷1011.7Molarity = 0.25
Profile image of Samyak Jain
7 Years ago
M(W + 1/m) = D, here M = Molarity of solution, W = Molar mass of solute in kg, m = Molality of solution,
                                    D = Density of solution.
D = 5/4 g/ml = 5/4 kg/L , m = 0.2 , W = molar mass of NaCl = (23 + 35.5)/1000 kg/mol = 0.0585 kg/mol
\therefore  M(0.0585 + 1/0.2) = 5/4
M(5.0585) = 5/4  \Rightarrow  M = 5 / (4 x 5.0585)  \approx  0.247