Question

Based upon the technique of reverse osmosis, the approximate minimum pressure required to desalinate sea water containing $3.25\%$ $\left(w/v\right)NaCl$ at ${27}^{\circ }\text{C}$ should be:
(Given: $R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}$)

• A. $27.33\text{atm}$
• B. $13.66\text{atm}$
• C. $2.73\text{atm}$
• D. $1.37\text{atm}$

Answer 1

The correct option is B $13.66\text{atm}$

Molecular mass of $NaCl=58.5\text{u}$
The pressure is $\pi =CRT=(\frac{3.25}{58.5}\times \frac{1000}{100})\times 0.082\times 300=13.66\text{atm}$