Question

0.85% aqueous solution of NaNO3 is apparantly 90% dissociated at 27.celcius

.calculate its osmotic pressure?

Answer 1

0.85 % of aq. solution of NaNO3 means 0.85 gm of NaNO3 is dissolved in 100 g of solution or 100 cm3 = 100ml of the solution.

Hence, here 0.85 gm of NaNO3 is dissolved in 100 ml of solution= 0.1L of solution,

Temperature = 27 °C = 273 +27 = 300 K

Now we know that osmotic pressure,

P = CRT = n/V xR xT

Where C = molar concentration

Now, no. of moles of NaNO3 in given sample = given mass/ molar mass = 0.85 / 85 (molar mass of NaNO3) = 0.01 moles.

So, P = n/V x R xT where, R = 0.0821 L.atm/mol.K

Putting the values we get the osmotic pressure as:

P = 0.01/0.1 x 0.0821 x 300 = 2.463 atm.

Answer 2

$\pi = crt \\ \pi = \frac{n}{v} rt \\ \pi = \frac{ \frac{0.85}{85} }{0.1} \times 0.082 \times 300 \\ \\ \pi = 2.46 \: atm$

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