Question

ionic strength (M) of 50 ml of 0.75 % (w/v) NaCl solution?

Answer 1

M = 0.75 M
V = 50 mL or 0.050 L
moles = P

 moles = P 

0.75M= P    =>mol  =0.050L

0.050L=P  = mol

0.0375 = P =mole

0.0375 mol    NaCl    P - 58.44g     

NaCl

1mol = NaCl

=2.19 grams NaCl

Answer 2

Answer : The ionic strength of solution is, 0.128 mole/L

Explanation :

First we have to calculate the mass of NaCl in 50 ml solution.

As, 100 ml of solution contains 0.75 grams of NaCl

So, 50 ml of solution contains $\frac{0.75}{100}\times 50=0.375g$ of NaCl

Now we have to calculate the concentration or molarity of solution.

$Molarity=\frac{\text{Mass of NaCl}\times 1000}{\text{Molar mass of NaCl}\times \text{volume of solution in ml}}=\frac{0.375g\times 1000}{58.5g/mole\times 50ml}=0.128mole/L$

Now we have to calculate the ionic strength of solution.

Formula used for ionic strength of solution :

$M=\frac{1}{2}\sum C\times Z^2$

where,

M = ionic strength of solution

C = concentration of solution  = 0.128 mole/L

Z = charge on ions in the solution

$M=\frac{1}{2}[C\times (\text{Charge on }Na^+)^2+C\times (\text{Charge on }Cl^-)^2]$

$M=\frac{1}{2}[0.128\times (+1)^2+0.128\times (-1)^2]=0.128mole/L$

Therefore, the ionic strength of solution is, 0.128 mole/L