Which of the following pair of solutions is isotonic ?

Question

Which of the following pair of solutions is isotonic ? (A) 0.01M BaCl2 and 0.015M NaCl (B) 0.001M Al2(SO4)3 and 0.01 M BaCl2 (C) 0.001M CaCl2 and 0.001M Al2(SO4)3 (D) 0.01M BaCl2 and 0.001M CaCl2

Tardigrade Admin · Accepted Answer

Isotonic solutions are those which have the same osmotic pressure

(π = i CRT)

. 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 C a C l_{2}, i = 3

.
So

i \times C = 3 \times 0.001 = 0.003

For,

0.001 M A l_{2} (S O_{4})_{3}, i = 5.

So

i \times C = 5 \times 0.001 = 0.005

b. For,

0.01 M B a C l_{2}, I = 3.

So

i \times C = 3 \times 0.01 = 0.03

For,

0.001 M C a C l_{2}, i = 3.

So

i \times C = 3 \times 0.001 = 0.003

c. For,

0.01 M B a C l_{2}, i = 3

.
So

i \times C = 3 \times 0.01 = 0.03

For,

0.015 M N a Cl, i = 2.

So

i \times C = 2 \times 0.015 = 0.03

Thus

0.01 M B a C l_{2}

and

0.015 MN a Cl

are isotonic in nature.

d. For,

0.001 M A l_{2} (S O_{4})_{3}, i = 5.

So

i \times C = 5 \times 0.001 = 0.005

For,

0.01 M a C l_{2}, i = 3.

So

i \times C = 3 \times 0.01 = 0.03

Q. Which of the following pair of solutions is isotonic ?

$0.01 M B a C l_{2}$ and $0.015 M N a Cl$

$0.001 M A l_{2} (S O_{4})_{3}$ and $0.01 M B a C l_{2}$

$0.001 M C a C l_{2}$ and $0.001 M A l_{2} (S O_{4})_{3}$

$0.01 M B a C l_{2}$ and $0.001 M C a C l_{2}$

Q. Which of the following pair of solutions is isotonic ?

0.01MBaCl2​ and 0.015MNaCl

0.001MAl2​(SO4​)3​ and 0.01MBaCl2​

0.001MCaCl2​ and 0.001MAl2​(SO4​)3​

0.01MBaCl2​ and 0.001MCaCl2​

$0.01 M B a C l_{2}$ and $0.015 M N a Cl$

$0.001 M A l_{2} (S O_{4})_{3}$ and $0.01 M B a C l_{2}$

$0.001 M C a C l_{2}$ and $0.001 M A l_{2} (S O_{4})_{3}$

$0.01 M B a C l_{2}$ and $0.001 M C a C l_{2}$