What volume, in millilitres, of distilled water must be added to 105 mL of 0.600 mol/L sulfuric acid solution in order to dilute it to a concentration of 0.400 mol/L
moles H2SO4 = M x L = 0.105*0.600 = ??
M = moles/L
You know moles and M, solve for L and convert to mL. Technically, this is the volume to which the H2SO4 must be diluted to make the final volume; however, the question asks for the amount of water that must be added. Therefore, assuming the volume are additive , you want the final volume you calculate -105 mL = amount to be added.
much appreciated
To calculate the volume of distilled water needed to dilute the sulfuric acid solution, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 is the initial concentration of the sulfuric acid solution
V1 is the initial volume of the sulfuric acid solution
C2 is the final concentration desired
V2 is the final volume desired (which includes the volume of water to be added)
Given:
C1 = 0.600 mol/L
V1 = 105 mL
C2 = 0.400 mol/L
Rearranging the formula to solve for V2, we have:
V2 = (C1V1) / C2
Substituting the given values:
V2 = (0.600 mol/L * 105 mL) / 0.400 mol/L
Calculating the result:
V2 = (63 mol * mL) / 0.4 mol
V2 = 157.5 mL
Therefore, you need to add 157.5 mL of distilled water to the 105 mL of sulfuric acid solution in order to dilute it to a concentration of 0.400 mol/L.
To dilute the sulfuric acid solution, you need to determine the volume of water to be added. The general formula for dilution is:
\(C_1V_1 = C_2V_2\)
where:
- \(C_1\) represents the initial concentration of the solution (0.600 mol/L)
- \(V_1\) represents the initial volume of the solution (105 mL)
- \(C_2\) represents the final concentration of the solution (0.400 mol/L)
- \(V_2\) represents the final volume of the solution (initial volume + volume of water added)
We can rearrange the formula to solve for \(V_2\) to determine the final volume of the diluted solution:
\(V_2 = \frac{{C_1 \cdot V_1}}{{C_2}}\)
Substituting the given values, we have:
\(V_2 = \frac{{0.600 \, \text{mol/L} \cdot 105 \, \text{mL}}}{{0.400 \, \text{mol/L}}}\)
\(V_2\) is the final volume of the diluted solution, which is equal to the initial volume (105 mL) plus the volume of water to be added. Therefore, we can write the equation as:
\(105 \, \text{mL} + \text{volume of water added} = \frac{{0.600 \, \text{mol/L} \cdot 105 \, \text{mL}}}{{0.400 \, \text{mol/L}}}\)
Now, let's solve for the volume of water added:
\(\text{volume of water added} = \frac{{0.600 \, \text{mol/L} \cdot 105 \, \text{mL}}}{{0.400 \, \text{mol/L}}} - 105 \, \text{mL}\)
Calculating this expression will give you the volume of water, in milliliters, that needs to be added to the sulfuric acid solution to dilute it to a concentration of 0.400 mol/L.