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Question

Chemistry Question on Colligative Properties

Which of the following pair of solutions is isotonic ?

A

0.01MBaCl20.01M \, BaCl_2 and 0.015MNaCl0.015M \, NaCl

B

0.001MAl2(SO4)30.001M \, Al_2(SO_4)_3 and 0.01MBaCl20.01 \,M \,BaCl_2

C

0.001MCaCl20.001M \, CaCl_2 and 0.001MAl2(SO4)30.001M \, Al_2(SO_4)_3

D

0.01MBaCl20.01M \,BaCl_2 and 0.001MCaCl20.001M \, CaCl_2

Answer

0.01MBaCl20.01M \, BaCl_2 and 0.015MNaCl0.015M \, NaCl

Explanation

Solution

Isotonic solutions are those which have the same osmotic pressure (π=iCRT)(\pi = iCRT). But here we have different concentration of the solutions and also they have different Van't Hoff factors (i). So the solutions for which the product of ii and cc will be the same and isotonic.

a. For, 0.001MCaCl2,i=30.001\, M \,CaCl _{2}, i =3.
So i×C=3×0.001=0.003i \times C =3 \times 0.001=0.003

For, 0.001MAl2(SO4)3,i=5.0.001 \,M\, Al _{2}\left( SO _{4}\right)_{3}, i =5 .
So i×C=5×0.001=0.005i \times C =5 \times 0.001=0.005

b. For, 0.01MBaCl2,I=3.0.01 \,M\, BaCl _{2}, I =3 .
So i×C=3×0.01=0.03i \times C =3 \times 0.01=0.03

For, 0.001MCaCl2,i=3.0.001\, M \,CaCl _{2}, i =3 .
So i×C=3×0.001=0.003i \times C =3 \times 0.001=0.003

c. For, 0.01MBaCl2,i=30.01 \,M \,BaCl _{2}, i =3.
So i×C=3×0.01=0.03i \times C =3 \times 0.01=0.03

For, 0.015MNaCl,i=2.0.015 \,M \,NaCl , i =2 .
So i×C=2×0.015=0.03i \times C =2 \times 0.015=0.03

Thus 0.01MBaCl20.01 \,M \,BaCl _{2} and 0.015MNaCl0.015 M NaCl are isotonic in nature.

d. For, 0.001MAl2(SO4)3,i=5.0.001 \,M\, Al _{2}\left( SO _{4}\right)_{3}, i =5 .
So i×C=5×0.001=0.005i\times C =5 \times 0.001=0.005

For, 0.01MaCl2,i=3.0.01 \,M \,aCl _{2}, i =3 .
So i×C=3×0.01=0.03i \times C =3 \times 0.01=0.03