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Question: Determine the amount of \(CaC{l_2}(i = 2.47)\) dissolved in \(2.5\) litre of water such that its osm...

Determine the amount of CaCl2(i=2.47)CaC{l_2}(i = 2.47) dissolved in 2.52.5 litre of water such that its osmotic pressure is 0.75 atm at 27C0.75{\text{ atm at 2}}{{\text{7}}^\circ }C.

Explanation

Solution

The osmotic pressure can be given as the product of van't Hoff’s factor(ii), molar concentration(CC), real gas constant (RR) and temperature(TT). Osmotic pressure can be expressed as Π=iCRT\Pi = iCRT.

Complete step by step answer:
In this question, osmotic pressure can be defined as the minimum pressure that must be applied to a solution to prevent the flow of solvent molecules through semipermeable membranes. Here, semipermeable membrane is a membrane that allows the movement of only solvent particles through it, semipermeable membrane does not allow solute particles to move through it .
Osmotic pressure is a colligative property because it depends on the concentration of solute particles in the solution. Osmotic pressure can be calculated using the Van't Hoff’s equation that can be given as:
Π=iCRT\Pi = iCRT (1) - (1)
Where Π=\Pi = osmotic pressure
i=i = van't Hoff’s factor
C=C = molecular concentration of the solute in solution
R=R = Universal gas constant
T=T = Temperature
Now, we are given in the question the following values:
i=2.47i = 2.47
V=2.5LV = 2.5L (Volume)
Π=0.75atm\Pi = 0.75atm
T=27+273=300KT = 27 + 273 = 300K (Here we converted celsius temperature into kelvin scale)
R=0.0812atmK1mol1R = 0.0812atm{K^{ - 1}}mo{l^{ - 1}}
Now, as we know that molar concentration (CC) can be given as:
C=nVC = \dfrac{n}{V} where n=n = number of moles, V=V = Volume
Now, putting above value in equation (1) - (1) we get:
Π=iRTnV\Pi = \dfrac{{iRTn}}{V}
Now, by putting all values in above equation we get,
0.75=2.47×0.0812×300×n2.50.75 = \dfrac{{2.47 \times 0.0812 \times 300 \times n}}{{2.5}}
n=0.0308n = 0.0308
As we know that number of moles is equal to given mass divided by molecular mass. Now, the molecular mass of calcium chloride is 111g/mol111g/mol.
Now, Amount of calcium chloride dissolved =111×0.0308=3.42g = 111 \times 0.0308 = 3.42g
Hence, 3.42g3.42g of calcium chloride is dissolved in water.

Note:
It is important to note that the Van't Hoff’s equation holds true only in the case of solutions that behave like ideal solutions hence, it is not applicable for the solutions that behave like non-ideal solutions.