Question

5.85gms of sodium chloride were dissolved in water and the solution made upto 100ml standard flask.10ml of

this solution were pipette out into another flask and made up with dissolved water into 100ml of solution. The

concentration if the sodium chloride solution now is



(A)0.1M

(B)1.0M

(D)0.5M

(D)0.25M


solve with explanation​

Answer 1

Answer:Mole =  

molecular weight

weight

​  

 

Molecular weight of NaCl is 58.5

molarity of NaCl =  

volume in L

mol

​  

 

                           =  

58.5×0.2

5.85

​  

  = 0.5M

Molecular weight of NaBr is 103.

molarity of NaBr =  

volume in L

mol

​  

 

                          =  

103×0.5

20.6

​  

 = 0.4M

Explanation:

Answer 2

The concentration if the sodium chloride solution now is 0.1M

Given:

Mass of NaCl = 5.85 gm

volume of the solution= 100 mL=0.1 L

Solution:

∴ No. of moles of NaCl in solution= $\frac{5.85}{58.5}$ gm = 0.1 [∵molar mass of NaCl = 58.5gm]

Molarity(M₁)= $\frac{No. of moles}{volume of solution}$  = $\frac{0.1}{0.1} M$ =1M

Now volume (V₁) of this solution taken= 10mL

The new volume(V₂)= 100mL

∴Let's take the concentration of new NaCl solution is= M₂ M

As we know, dilution formula,

   M₁V₁ = M₂V₂

⇒ 1×10 = M₂ × 100

⇒M₂= 0.1

∴ The concentration if the sodium chloride solution now is 0.1M.

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