What volume of water is required to dilute 120 cm3 of 10 mol dm–3 sulphuric acid to a concentration of 2 mol dm–3?

Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 sulphuric acid to a concentration of 2 .

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

In this case, you know:

Replacing in the definition of dilution:

10× 120 cm³= 2 × Vf

Solving:

Vf= (10× 120 cm³) ÷2

Vf= 600 cm³

In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 sulphuric acid to a concentration of 2 .

Learn more about dilution:

0.48 dm3  (or 480 cm3)

Explanation:

First find the original no. of moles existing in the sulphuric acid:

no. of moles = volume (in dm3) x concentration

                     = 120/1000 x 10

                     = 1.2 mol

Then let the total volume of the diluted acid be v dm3.

Since

Concentration = no. of moles / volume,

so by substituting the given information,

2 = 1.2 / v

v = 0.6 dm3

Hence, the volume of water required

= 0.6 - 120/1000

= 0.48 dm3  (or 480 cm3)

the answer is d   for sure : )

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it