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.
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.
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Put the value of R=8.314 instead of putting 0.0821 ,then will get the right answer that is 0.03mol.
But sir NCERT me to answer 0.03 mol de rha hai Is answer se to 0.003 mol aa rha hai Kya karu
Good site.Has each answer.
I help this answer