# Need to calculate the resistance accross membrane in water

Hello everybody,
I am chemist. and I prepare ionic membrane and placed in a home made acrylic cell containing 0.1M NaCl separated into two portions by ionic membrane. I applied DC voltage of 9.6V (using 9V battery). Using multi-meter the measured DC voltage in water was 4.45V. The distance between the electrodes is 20cm. The measured amperes were 0.93mA. Can I calculate the resistance from voltage drop? and how? Please help or guide if I m doing wrong.

## Answers and Replies

It's not clear to me what you are doing, but a diagram might help. My best guess is that the resistance is 4.45V divided by 0.93 mA which is 4785 Ohms. But without an understanding of what you are doing, that caculcation might not be the right one, or the experimental method could be fundamentally flawed.

The thing that seems strange is the battery is loaded down to 4.45 V with less than 1 mA of current. Not sure why this is.

jim hardy
Science Advisor
Gold Member
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Using multi-meter the measured DC voltage in water was 4.45V.
Voltage is always between two points.
"voltage in water" is just three words that convey no information.

Where were the two voltmeter leads physically placed ?