Question

A 0.300M solution of HCl is prepared by adding some 1.50M HCl to a 500mL volumetric flask and diluting to the mark with deionized water . What volume of 1.50M HCl must be added ?

Answer 1

No. of moles of HCl = volume in litres x molarity
                                = (500/1000) x 0.3 
                                = 0.15 moles

No. of moles og HCl in the concentrated solution = 0.15 moles
No. of moles of HCl = volume in litres x molarity
0.15 = volume in litres x 1.5
Volume in litres = 0.15/1.5 = 0.1 Litres
Volume = 100 ml

Answer 2

Hello Dear.

Here is the answer ⇒

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Given ⇒

In First Case,

Molarity of the HCl = 0.300 M.
Volume of the HCl(in ml) = 500 ml.

Using the Formula,

 Molarity = (No. of moles/Volume of the HCl in ml) × 1000.

⇒ No. of Moles =  (Molarity × Volume of the HCl)/1000.
⇒ No. of moles = 0.3 × 500/1000.
                         = 0.15 moles.


In Second Case,

 Molarity of HCl which is added = 1.50 M.

No. of moles of the HCl = 0.15 moles
        [Calculated above]

∴ Using the Formula,

  Molarity = (No. of moles/Volume of the HCl in ml) × 1000

⇒ 1.5 =   (0.15/Volume of the HCl) × 1000
⇒ Volume of the HCl = 150/1.5 
⇒ Volume of the HCl in ml = 100 ml.


Thus, the Volume of the HCl of Molarity 1.5 M which is added to the HCl of Molarity 0.3 M is 100 ml.


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Hope it helps.

Have a Good Day.