Question

At body temperature (37°C), blood has an osmotic pressure of about 8.2atm. To match this pressure with NaCl solution (saline), how many grams of NaCl should you dissolve in water for 1.00L of solution?

Answer 1

first find the molarity using

= M x R x T x i

where

= osmatic pressure = 8.2 atm

M = Molarity we have to calulate

R = gas constant = 0.0821 L atm / mol K

T = temperature in kelvin = 37 ºC = 273 + 37 = 310 K

i = vant haf factor which is 2 for naCl

plug in these values in above equation

8.2 atm = M x 0.0821 L atm / mol K x 310 K x 2

M = 8.2 atm / 50.902 atm L / mol

M = 0.1611 mol /L

now

Molarity = no fo moles / volume in liters

volume is given as 1L so

0.1611 = mol / 1L

mol = 0.1611

Mass = moles x molar mass = 0.1611 mol x 58.443 g/mol = 9.4 grams

9.4 grams NaCl