What mass of MgSO4 · 7H2O is required to prepare 800 mL of a 0.442MMgSO4 solution?
Answer in units of g.
M = mol/V
0.442(mol/L)= mol/0.8(L)
MOL = 0.3536 mol
MASS= 0.3536 * Mm MgSO4.7H2O
Complete and answer the following equation
___Mg(ClO3)2 + ___(NH4)2SO4 ---->
How to balance and figure out what type of chemical reaction.
To find the mass of MgSO4 · 7H2O required to prepare the solution, we need to use the formula:
Molarity = Moles of solute / Volume of solution (in liters)
First, let's convert the volume of the solution from milliliters to liters:
800 mL = 800/1000 = 0.8 L
Next, we rearrange the formula to solve for Moles of solute:
Moles of solute = Molarity x Volume of solution
Substituting the given values:
Moles of solute = 0.442 M x 0.8 L = 0.3536 moles
Now, we'll use the molar mass of MgSO4 · 7H2O to convert moles to grams. The molar mass of MgSO4 · 7H2O can be calculated by summing up the atomic masses of its constituent elements:
Mg: 24.31 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (there are four oxygen atoms)
H: 1.01 g/mol (there are 14 hydrogen atoms)
Molar mass of MgSO4 · 7H2O = (24.31 g/mol) + (32.07 g/mol) + (4 x 16.00 g/mol) + (7 x 1.01 g/mol) = 246.47 g/mol
Now we can calculate the mass of MgSO4 · 7H2O:
Mass = Moles x Molar mass
Mass = 0.3536 moles x 246.47 g/mol = 87.25 grams
Therefore, 87.25 g of MgSO4 · 7H2O is required to prepare 800 mL of a 0.442 M MgSO4 solution.