A 1.17 % solution of NaCl is isotonic with 7.2 % solution of glucose. Calculate the van’t Hoff factor of NaCl

Question

A 1.17 % solution of NaCl is isotonic with 7.2 % solution of glucose. Calculate the van’t Hoff factor of NaCl (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For isotonic solution osmotic pressure should be the same ∴ π(NaCl) =π text(glucose) iC1RT=C2 RT iC1=C2 i ((1.17/ 58.5)/V)=(7.2 /180/V) ⇒ i=(7.2×58.5/1.17×180)=2

Q. A $1.17%$ solution of $N a Cl$ is isotonic with $7.2%$ solution of glucose. Calculate the van’t Hoff factor of $N a Cl$

For isotonic solution osmotic pressure should be the same
$∴ π_{(N a Cl)} = π_{(glucose)}$
$i C_{1} RT = C_{2} RT$
$i C_{1} = C_{2}$
$i \frac{( 1.17/58.5 )}{V} = \frac{7.2/180}{V}$
$\Rightarrow i = \frac{7.2 \times 58.5}{1.17 \times 180} = 2$

Q. A 1.17% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the van’t Hoff factor of NaCl

For isotonic solution osmotic pressure should be the same ∴π(NaCl)​=π(glucose)​ iC1​RT=C2​RT iC1​=C2​ iV(1.17/58.5)​=V7.2/180​ ⇒i=1.17×1807.2×58.5​=2

Q. A $1.17%$ solution of $N a Cl$ is isotonic with $7.2%$ solution of glucose. Calculate the van’t Hoff factor of $N a Cl$

For isotonic solution osmotic pressure should be the same
$∴ π_{(N a Cl)} = π_{(glucose)}$
$i C_{1} RT = C_{2} RT$
$i C_{1} = C_{2}$
$i \frac{( 1.17/58.5 )}{V} = \frac{7.2/180}{V}$
$\Rightarrow i = \frac{7.2 \times 58.5}{1.17 \times 180} = 2$