What is the osmotic pressure of 0.9% m/v solution of NaCl at 25°C?
Osmotic pressure "= i \\times C \\times R \\times T"
where, C= concentration of solute(in terms of Molarity)
"R = 0.082"
"i = 1"
0.9% "NaCl" solution means 0.9g is present in 100mL of solution.
Moles of "NaCl" "= \\dfrac{0.9}{58} = 0.015"
Concentration of "NaCl = \\dfrac{0.015}{100} \\times 1000 = 0.10"
Hence, Osmotic pressure "= 1 \\times 0.10 \\times 0.082 \\times 300"
"= 2.46atm"
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