Question

The diffusion coefficient for sodium ions crossing a biological membrane 12nm thick is 1.5 x10-18 m2/s. what flow rate of sodium ions would move across an area 12nm x 12 nm if the concentration difference across the membrane is 0.60 mol/dm3

Answer 1

`-1.08* 10^(23) mol/s`

Step-by-step explanation:

We are given that

Diffusion coefficient,
`D=1.5* 10^(-18) m^2/s`

Thickness of membrane,
`dx=12nm=12* 10^(-9) m`

`1 nm=10^(-9) m`

Area,
`A=12* 12=144nm^2=144* 10^(-18) m^2`

Concentration differences,
`dc=0.60 mol/dm^3=0.60* 1000=600mol/m^3`

We have to find the flow rate of sodium ions.

Flow rate,
`(dn)/(dt)=-DA(dc)/(dx)`

Using the formula

`(dn)/(dt)=-(1.5* 10^(-18)* (144* 10^(-18))* 600)/(12* 10^(-9))=-1.08* 10^(23) mol/s`

`(dn)/(dt)=-1.08* 10^(23) mol/s`