Concentrated solution that it would take to prepare 2.65 l of 0.515 m Hcl upon dilution with water is 131 ml.
A solution that has a relatively large amount of dissolved solute is known a concentrated solution .
As we know,
And , moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
Now, V₁ = V₂ * M₂/M₁
Given,
M₂ = 0.515 M
V₂ =2.65 L
As the hydrochloric acid can be concentrated up to 38% p/V ,
Therefore, maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
Now, min V₁ = V₂ * M₂/ max M₁
= 2.65 L* 0.515 M/ 10.41 M
= 0.131 L
= 131 ml
So, the quantity required can go from 131 ml up to 2900 ml ( if M₁ = M₂)
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