Question

Calculate the osmotic pressure of a solution containing 17.1 g of cane sugar in 500 g of water at 300 k density of solution 1.034 g/ cm3

Answer 1

Answer:

2.52 atm

Explanation:

cane sugar is sucrose. Sucrose has molecular mass 342 g.

Number of moles of 17.1 g of cane sugar is 17.1/342=0.05 mol

Volume of the solution can be calculated by

$Density=\frac{mass}{volume}$

$1034 g/L=\frac{500 g}{Volume}$

$Volume = 0.483 L$

osmotic pressure = Conc*R*T

Concentration is calculated by number of moles / volume i.e. 0.05 / 0.483

Gas constant = 0.0821 $L.atm.K^{-1}mol^{-1}$

Osmotic pressure = 0.1035*0.0821*300

= 2.52 atm

Answer 2

Answer:

249.42 Pa

Explanation:

Osmotic pressure is given by following mathematical formula

π = cRT

Where

π = osmotic pressure

c = concentration

R = Universal gas constant = 8.314 J/mol - K

T = temperature

Molar mass of sugar = 342 grams

Given mass of sugar =  17.1 g

Number of moles of sugar = 17.1/342 = 0.05 moles

Mass of solvent (water) = 500 g

Mass of solution = mass of solvent + mass of solute

                            = 500 g + 17.1 g

                            = 517.1 g

Density of solution = 1.034 g/cm³

Volume of solution = mass/density

                                = 517.1/1.034

                                = 500 cm³ = 500 ml = 0.5 Liter

Molarity = Number of moles of solute/Volume of solution in Liter

              = 0.05/0.5

               = 0.1 M

T = 300 K

Now Putting numerical values in above mathematical formula

π = 0.1 x 8.314 x 300

   = 249.42 Pa