Question

Which of the following pair of solutions is isotonic ?

• A. $0.01M \, BaCl_2$ and $0.015M \, NaCl$
• B. $0.001M \, Al_2(SO_4)_3$ and $0.01 \,M \,BaCl_2$
• C. $0.001M \, CaCl_2$ and $0.001M \, Al_2(SO_4)_3$
• D. $0.01M \,BaCl_2$ and $0.001M \, CaCl_2$

Answer 1

The Correct Option is A

Isotonic solutions are those which have the same osmotic pressure $(\pi = iCRT)$. 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 $i$ and $c$ will be the same and isotonic.

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

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

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

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

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

For, $0.015 \,M \,NaCl , i =2 .$
So $i \times C =2 \times 0.015=0.03$

Thus $0.01 \,M \,BaCl _{2}$ and $0.015 M NaCl$ are isotonic in nature.

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

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