Question

Sea water is $3.5$% by mass of common salt and has a density $1.04g$ ${cm}^{-3}$ at $298K$. Assuming the salt to be sodium chloride, then osmotic pressure of sea water will be: (assume complete ionisation of the salt)

• A. $25.45atm$
• B. $11.56atm$
• C. $15.13atm$
• D. $30.40atm$

Answer 1

Answer: (D)

$30.40atm$

Given that:
$\%bymassofNaCl=3.5\%$ in sea water

So in 100 g of sea water, Mass of NaCl=3.5g

Molecular weight of NaCl = 58.5 g

No. of moles = $\frac{3.5}{58.5}$

Density of sea water$=1.04gc{m}^{-3}$ at T=298 K
$V=m/d$

Volume of solution = $\frac{100g}{(1.04\times 1000)g/L}$

Molarity, $M=\frac{3.5}{58.5}\times \frac{1.04\times 1000}{100}$

Gas constant, R = 0.0821

Temperature, T = 298 K

Vant hoff factor, i = 2

Thus osmotic pressure will be,

$\pi =iMRT$

$\pi =2\times \frac{3.5}{58.5}\times \frac{1.04\times 1000}{100}\times 0.0821\times 298$

$\pi =30.40atm$

Correct option is D.