This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Answer:
0.071L
Explanation:
From the question given, we obtained the following data:
Molarity of HCl = 2.25 M
Mass of HCl = 5.80g
Molar Mass of HCl = 36.45g/mol
Number of mole of HCl =?
Number of mole = Mass /Molar Mass
Number of mole of HCl = 5.8/36.45 = 0.159mole
Now, we can obtain the volume required as follows:
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.159mole/ 2.25
Volume = 0.071L
Answer:
0.00268 M
Explanation:
To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.
<u>Step 1:</u>
3.00 mL / 1,000 = 0.00300 L
Molarity = moles / volume (L)
0.0250 M = moles / 0.00300 L
(0.0250 M) x (0.00300 L) = moles
7.50 x 10⁻⁵ = moles
<u>Step 2:</u>
25.0 mL / 1,000 = 0.0250 L
0.0250 L + 0.00300 L = 0.0280 L
Molarity = moles / volume (L)
Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)
Molarity = 0.00268 M
Answer:
6.53g of K₂SO₄
Explanation:
Formula of the compound is K₂SO₄
Given parameters:
Volume of K₂SO₄ = 250mL = 250 x 10⁻³L
= 0.25L
Concentration of K₂SO₄ = 0.15M or 0. 15mol/L
Unknown:
Mass of K₂SO₄ =?
Methods:
We use the mole concept to solve this kind of problem.
>>First, we find the number of moles using the expression below:
Number of moles= concentration x volume
Solving for number of moles:
Number of moles = 0.25 x 01.5
= 0.0375mole
>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:
Mass(g) = number of moles x molar mass
Solving:
To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.
For:
K = 39g
S = 32g
O = 16g
Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)
= 78 +32 + 64
= 174g/mol
Using the expression:
Mass(g) = number of moles x molar mass
Mass of K₂SO₄ = 0.0375 x 174 = 6.53g
