a 340.4 ml sample of solution contains 27.4 g of cacl2. calcuate the molar concentration of calcium ion in this solution.
this is what i did
27.4(110.98) =.247mol
.247mol/34.4L=.726Mcacl2
is this correct?
ARRRRGGGGG.
moles=27.4g/(110.98) =.247mol
molarity=.247/.3404liters=.726Moles/liter
whats arrgg mean?
To calculate the molar concentration of calcium ions in the solution, you first need to determine the amount of calcium ions present in the solution.
Given:
Volume of solution (V) = 340.4 mL = 0.3404 L
Mass of CaCl2 (m) = 27.4 g
To find the amount of calcium ions (moles), you need to convert the mass of CaCl2 to moles using its molar mass (110.98 g/mol).
Amount (moles) of CaCl2 = mass (g) / molar mass (g/mol)
Amount (moles) of CaCl2 = 27.4 g / 110.98 g/mol
Amount (moles) of CaCl2 = 0.2468 mol
Now, in one mole of CaCl2, there is one mole of calcium ions (Ca^2+) and two moles of chloride ions (Cl^-).
Since we are interested in the molar concentration of calcium ions, we can say that the moles of calcium ions (Ca^2+) is equal to the moles of CaCl2.
Now we have the amount of calcium ions (0.2468 mol) in the solution. However, we need to express this in terms of concentration, which is moles per liter (M).
Molar concentration (M) = amount of solute (moles) / volume of solution (liters)
Molar concentration of calcium ions (Ca^2+) = 0.2468 mol / 0.3404 L
Molar concentration of calcium ions (Ca^2+) = 0.724 M
Therefore, the correct molar concentration of calcium ions (Ca^2+) in the solution is 0.724 M.
So, your calculation was close, but it seems you made a small arithmetic mistake in the division. It should be 0.2468 mol / 0.3404 L, which equals 0.724 M.