Question

What would be the osmotic pressure at 25°C of an aqueous solution containing 1.95g of sucrose present in 150 ml of solution?

Answer 1

Hi mate.

Thanks for asking this question,

Here is your answer,


Given :

T = 25c = 298k

V = 150 ml.

n = w/m = 1.95 / 342 = 0.005 moles.

R = 8.314 J/k.mol.


$\pi \: =$

We know that,


$\pi \: v \: = \: n \: r \: t$
Therefore,


$\pi \ \times 150 \: = \: 0.005 \times 8.314 \times 298$

$\pi \: \times 150 = \: 5 \times {10}^{? - 3 } \times 8.314 \times \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 298$

$\pi \: = \: 0.33 \times {10}^{? - 2} \times 2477.5$

$\pi \: = \: 3.3 \times {10}^{? - 3} \times 2477.5$

$\pi \: = 8175.98 \: \times {10}^{ - 3} pa.$








BE BRAINLY !!!!!!!!!!!

Answer 2

Answer: The osmotic pressure of the solution is 0.928 atm.

Explanation:

Osmotic pressure is calculated by using the formula:

$\pi=CRT\\\\\pi=\frac{n}{V}RT$     .....(1)

where,

$\pi$ = Osmotic pressure

n = number of moles

V = volume of solution ( in L)

R = Gas constant

T = Temperature of the solution

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sucrose = 1.95 g

Molar mass of sucrose = 342.3 g/mol

Putting values in above equation, we get:

$\text{Moles of sucrose}=\frac{1.95g}{342.3g/mol}\\\\\text{Moles of sucrose}=0.00569mol$

We are given:

$n=0.00569mol\\V=0.15L\text{ (Conversion factor: 1L=1000mL)}\\T=25^oC=298K\text{ (Conversion factor: }T(K)=T(^oC)+273)}\\R=0.0821\text{ L atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$\pi=\frac{0.00569mol}{0.15L}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\\pi=0.928atm$

Hence, the osmotic pressure of the solution is 0.928 atm.