_____ g of MnSO4 H2O are needed to prepare a solution of 100.00 mL of 0.700 M MnSO4 Dilution of 1.00mL of 0.7 M MnSO4 solution to 20.00 mL aqueous solution gives a new concentration of _____ M .
You want 0.700 M MnSO4 so you want 0.700 moles MnSO4 in 1 L of soln. For 100 mL, you will want 0.1 of that. You want to know grams and moles = grams/molar mass. You know moles and you know molar mass; calculate grams.
Then on the dilution, the easy way is to do it this way.
0.7 M x 1.00/20.00 = ?? but sometimes the easiest to understand is to change 0.7 M x 0.001 L (that's 1 mL) to moles, then put that number of moles in 20 mL, divide moles/L to convert to M. You should get the same answer.
To calculate the grams of MnSO4 H2O needed to prepare a solution of 100.00 mL of 0.700 M MnSO4, you need to use the formula:
grams = (moles) x (molar mass)
First, let's determine the moles of MnSO4 needed:
moles = (volume in liters) x (molarity)
= 0.100 L x 0.700 mol/L
= 0.070 mol
Next, we need to find the molar mass of MnSO4 H2O:
Molar mass of Mn = 54.938045 g/mol
Molar mass of S = 32.06 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H2O = 2(1.00784 g/mol) + 16.00 g/mol = 18.02 g/mol
Molar mass of MnSO4 H2O = (molar mass of Mn) + (molar mass of S) + 4 x (molar mass of O) + (molar mass of H2O)
= 54.938045 g/mol + 32.06 g/mol + 4 x 16.00 g/mol + 18.02 g/mol
= 169.92 g/mol
Finally, we can calculate the grams of MnSO4 H2O needed:
grams = (moles) x (molar mass)
= 0.070 mol x 169.92 g/mol
= 11.8944 g of MnSO4 H2O
So, approximately 11.89 g of MnSO4 H2O are needed to prepare a solution of 100.00 mL of 0.700 M MnSO4.
To calculate the new concentration (M) after dilution of 1.00 mL of 0.7 M MnSO4 solution to 20.00 mL, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
Plugging in the values, we have:
M1 = 0.7 M
V1 = 1.00 mL = 0.001 L
M2 = unknown
V2 = 20.00 mL = 0.020 L
Using the equation, we can solve for M2:
(0.7 M)(0.001 L) = (M2)(0.020 L)
0.0007 mol = 0.020 M2
M2 = 0.0007 mol / 0.020 L
M2 = 0.035 M
Therefore, the new concentration after dilution is 0.035 M.
To find the mass of MnSO4 H2O needed to prepare a solution of 100.00 mL of 0.700 M MnSO4, we need to use the formula:
Molarity (M) = (moles of solute) / (liters of solution)
First, we can calculate the number of moles of MnSO4 needed for the desired final volume:
0.700 M = (moles of MnSO4) / 0.100 L
Rearranging the equation:
moles of MnSO4 = 0.700 M * 0.100 L
moles of MnSO4 = 0.0700 mol
Next, we need to find the molar mass of MnSO4 H2O. The molar mass of MnSO4 is 151.0 g/mol, and the molar mass of H2O is 18.0 g/mol.
Molar mass of MnSO4 H2O = (151.0 g/mol) + (18.0 g/mol) = 169.0 g/mol
Finally, we can calculate the mass of MnSO4 H2O needed:
Mass of MnSO4 H2O = moles of MnSO4 * molar mass of MnSO4 H2O
Mass of MnSO4 H2O = 0.0700 mol * 169.0 g/mol
Mass of MnSO4 H2O = 11.830 g
Therefore, 11.830 grams of MnSO4 H2O are needed to prepare a solution of 100.00 mL of 0.700 M MnSO4.
Now, let's move on to the second question: Dilution of 1.00mL of 0.7 M MnSO4 solution to 20.00 mL aqueous solution gives a new concentration of _____ M.
To find the new concentration after dilution, we can use the formula:
M1V1 = M2V2
Where:
M1 is the initial concentration,
V1 is the initial volume,
M2 is the final concentration, and
V2 is the final volume.
Given:
M1 = 0.7 M
V1 = 1.00 mL
V2 = 20.00 mL
Let's substitute these values into the formula:
(0.7 M) * (1.00 mL) = (M2) * (20.00 mL)
Rearranging the equation to solve for M2:
M2 = (0.7 M * 1.00 mL) / 20.00 mL
M2 = 0.035 M
Therefore, the new concentration after dilution is 0.035 M.