Q. 18 glucose $\left(C_{6} H_{12} O_{6}\right)$ is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is
NTA AbhyasNTA Abhyas 2022
Solution:
Relative lowering of vapour pressure
Equation $\frac{P^{0} - P_{s}}{P^{0}}=X_{s o l u t e}=\frac{n}{n + N}$
Modified forms of equation is $\frac{P^{0} - P_{s}}{P_{s}}=\frac{n}{N}$
n = moles of solute, N = moles of solvent
$P^{0}=760$ torr
$P_{s}=?$
$\frac{760 - P_{s}}{P_{s}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}$
$P_{s}=752.4$ torr
Equation $\frac{P^{0} - P_{s}}{P^{0}}=X_{s o l u t e}=\frac{n}{n + N}$
Modified forms of equation is $\frac{P^{0} - P_{s}}{P_{s}}=\frac{n}{N}$
n = moles of solute, N = moles of solvent
$P^{0}=760$ torr
$P_{s}=?$
$\frac{760 - P_{s}}{P_{s}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}$
$P_{s}=752.4$ torr