Question

A tube of uniform cross-sectional area $1c{m}^2$ is closed at one end with the semipermeable membrane. a solution of 5 g glucose per 100 mL is placed inside the tube and is dipped in pure water at ${27}^{o}C$. When equilibrium is established the height developed in a vertical column in cm is 1000X. (Assume the density of final glucose solution 1g/mL)

Find the value of X.

Answer 1

$\because d={10}^{3}kg/m^3$

$g=9.81m/se{c}^2$

$\pi =hdg$

$6.842\times 1.01\times {10}^5=h\times {10}^3\times 9.81$

$h=\frac{6.842\times 1.01\times {10}^5}{9.81\times {10}^3}$

$=70.44m=7044cm=1000X$

so $X\approx 7$