How many moles of magnesium sulfate are needed to make in a solution of 0.10g Mg2SO4?
The question is worded poorly. I don't understand the question.
To determine the number of moles of magnesium sulfate needed, we first need to calculate the molar mass of Mg2SO4.
The molar mass of an element or compound is the mass of one mole of that substance. It is obtained by summing the atomic masses of each element present, as indicated by the chemical formula.
The atomic masses of the elements are as follows:
Magnesium (Mg): 24.31 g/mol
Sulfur (S): 32.07 g/mol
Oxygen (O): 16.00 g/mol
The chemical formula for magnesium sulfate is Mg2SO4, which indicates that there are two magnesium atoms (2 x Mg), one sulfur atom, and four oxygen atoms (4 x O) in one formula unit.
Now we can calculate the molar mass of Mg2SO4:
Molar mass of Mg2SO4 = (2 x Mg) + S + (4 x O)
= (2 x 24.31 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
= 48.62 g/mol + 32.07 g/mol + 64.00 g/mol
= 144.69 g/mol
The molar mass of magnesium sulfate is approximately 144.69 g/mol.
Next, let's calculate the number of moles of magnesium sulfate using the given mass:
Given mass of Mg2SO4 = 0.10 g
To calculate the number of moles, we can use the formula:
Number of moles = Given mass / Molar mass
Number of moles of Mg2SO4 = 0.10 g / 144.69 g/mol
= 0.000690 mol
Therefore, approximately 0.000690 moles of magnesium sulfate are needed to make a solution with a mass of 0.10 g of Mg2SO4.