what volume, in L, of a 10.0M HCL solution is needed to make 2.00L of 2.00m HCL solution by dilution with water.
Thank you
Why did the acid go to therapy? Because it had some serious concentration issues! Now, let's figure out the volume of that 10.0M HCL solution you need.
We can use the dilution equation, which states that the initial concentration multiplied by the initial volume is equal to the final concentration multiplied by the final volume. So, plug in the given values:
(10.0 M)(x L) = (2.00 M)(2.00 L)
Solving for x, we have:
x L = (2.00 M)(2.00 L) / (10.0 M)
x L = 0.40 L
So, you'll need 0.40 L (or 400 mL) of the 10.0M HCL solution to make 2.00 L of the desired 2.00M HCL solution. Now, be careful not to mix them up and create a truly explosive situation!
To solve this problem, we can use the dilution formula:
\( M_1V_1 = M_2V_2 \)
Where:
\( M_1 \) is the initial concentration of the solution (in this case, 10.0 M),
\( V_1 \) is the initial volume of the solution (unknown in this case),
\( M_2 \) is the final concentration of the solution (2.00 M),
\( V_2 \) is the final volume of the solution (2.00 L).
To find the volume of the initial 10.0 M HCl solution needed, we can rearrange the formula as:
\( V_1 = \frac{{M_2V_2}}{{M_1}} \)
Substituting the given values:
\( V_1 = \frac{{2.00 \, \text{M} \times 2.00 \, \text{L}}}{{10.0 \, \text{M}}} \)
Simplifying:
\( V_1 = \frac{{4.00 \, \text{mol} \cdot \text{L}}}{{10.0}} \)
\( V_1 = 0.40 \, \text{L} \)
Thus, you would need 0.40 L of the 10.0 M HCl solution to make 2.00 L of a 2.00 M HCl solution by dilution with water.
sorry
V1=C1/C2V2
C1V1=C2V2
C1= 10.0M
V1=?
C2= 2.00M
V2=2.00L
V2= C1/C2V2
The dilution formula is
c1v1 = c2v2
10x = 2*2
Solve for x.