Question

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Answer 1

We know that

π = i n/V RT

⇒π = i w/MV iRT

⇒ w = πMV /  iRT     .......................(1)

Now we have given below values:

π =  0.75 atm

V = 2.5L

i = 2.47

T = (27+273) K = 300K

Here,

R = 0.0821L atm k-1 mol-1

M = 1x40 + 2x35.5

= 111 g/mol

Now putting the value in equation 1:

w =  0.75x111x2.5 / 2.47x0.0821x300

=3.42g

Hence, the required amount of CaCl2 is 3.42 g.

Answer 2

Hey dear!!❤!!

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