0.004 M Na2SO4 is isotonic with 0.01 M glucose. Degree of dissociation of Na2SO4 is

Question

0.004 M Na2SO4 is isotonic with 0.01 M glucose. Degree of dissociation of Na2SO4 is (A) 75% (B) 50% (C) 25% (D) 85%

Tardigrade Admin · Accepted Answer

For isotonic solutions, they must have same concentrations of ions, Therefore, 0.004 i( Na 2 SO 4) =0.01 ⇒ i =(0.01/0.004)=2.5 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/45a0bbc8a7004c0435d909ab0cd85a97-.png" /> i =1+2 α=2.5 α=0.75 or 75 % dissociation

Q. $0.004 M N a_{2} S O_{4}$ is isotonic with $0.01 M$ glucose. Degree of dissociation of $N a_{2} S O_{4}$ is

75%

50%

25%

85%

For isotonic solutions, they must have same concentrations of ions, Therefore,
$0.004 i (N a_{2} S O_{4}) = 0.01$
$\Rightarrow i = \frac{0.01}{0.004} = 2.5$

$i = 1 + 2 α = 2.5$
$α = 0.75$
or $75%$ dissociation

Q. 0.004MNa2​SO4​ is isotonic with 0.01M glucose. Degree of dissociation of Na2​SO4​ is

75%

50%

25%

85%

For isotonic solutions, they must have same concentrations of ions, Therefore, 0.004i(Na2​SO4​)=0.01 ⇒i=0.0040.01​=2.5 i=1+2α=2.5 α=0.75 or 75% dissociation

Q. $0.004 M N a_{2} S O_{4}$ is isotonic with $0.01 M$ glucose. Degree of dissociation of $N a_{2} S O_{4}$ is

For isotonic solutions, they must have same concentrations of ions, Therefore,
$0.004 i (N a_{2} S O_{4}) = 0.01$
$\Rightarrow i = \frac{0.01}{0.004} = 2.5$

$i = 1 + 2 α = 2.5$
$α = 0.75$
or $75%$ dissociation