Question

Calculate osmotic pressure of $0.585$% $NaCl$ solution at ${27}^{0}C$.

Answer 1

Osmotic pressure $=i\times C\times R\times T,$

where, C=concentration of solute(in terms of Molarity)

R= Gas constant$=0.082L\left(atm\right)\left(mol\right)-1K-1$

T=temperature (in Kelvin)

i=Van’t-Hoff factor(=1 for non-electrolyte)

$0.585\%$ NaCl solution means $0.585g$ Nacl is present in $100ml$ of solution.

mole of NaCl is equal to the weight given is divided by the Molecular weight of NaCl,

$=\frac{0.585g}{58gmol-1}=0.01$

Thus, Concentration(C) of NaCl (in terms of Molarity)

$=\frac{moleofNaCl\left(n\right)}{Volumeofsolution}\times 1000$

$=\frac{\left(0.01\right)}{100}\times 1000$

$=0.10$

Hence Osmotic pressure $=1\times \left(0.10\right)\times 0.082\times 300atm$

$=2.46atm$