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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In First Case,

Molarity of the HCl = 0.3 M.
Volume of the HCl = 500 ml.

Using the Formula,

Molarity = (No. of moles/Volume in ml) × 1000
⇒ No. of moles = 0.3 × 500/1000
⇒ No. of moles = 0.15 moles.

In Second Case,

Molarity of the HCl = 1.5 M.
No. of moles = 0.15 moles
 [Calculate above]

Again using the same Formula which is used above,

Volume of the HCl in ml = (No. of moles/Molarity) × 1000
                                      = (0.15/1.5) × 1000
                                      = 150/1.5
                                      = 100 ml.

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


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Hope it helps.

Have a Good Day.