How many grams of #MgSO_4# are needed to prepare 125mL 0.200M magnesium sulfate?

Question

Luna Chen · Accepted Answer

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#

And #"mass of solute "=" moles"xx"molar mass"#

And so we calculate the product:

#"Volume"xx"Concentration"xx"Molar mass of magnesium sulfate"#

#=125xx10^-3cancel(L)xx0.200*cancel(mol*L^-1)xx120.37*g*cancel(mol^-1)#, an answer with units of mass as required.

#=??*g#

Cameron Andrews · Answer

undefined To calculate the grams of MgSO4 needed, you can use the formula:

Mass (g) = Volume (L) × Concentration (mol/L) × Molar mass (g/mol)

Given:
Volume = 125 mL = 0.125 L
Concentration = 0.200 M
Molar mass of MgSO4 = 120.37 g/mol

Plugging in the values:

Mass (g) = 0.125 L × 0.200 mol/L × 120.37 g/mol
Mass (g) = 3.005 g

So, you would need 3.005 grams of MgSO4 to prepare 125 mL of 0.200 M magnesium sulfate solution.

Wyatt Adkins · Answer

undefined To find the number of grams of MgSO4 needed to prepare 125 mL of 0.200 M magnesium sulfate solution, we use the formula:

$$ \text{moles} = \text{molarity} \times \text{volume (in liters)} $$

$$ \text{mass (in grams)} = \text{moles} \times \text{molar mass} $$

First, we calculate the moles of MgSO4:

$$ \text{moles} = 0.200 \, \text{M} \times 0.125 \, \text{L} = 0.025 \, \text{moles} $$

Then, we find the molar mass of MgSO4:

$$ \text{Molar mass of MgSO4} = 24.305 + 32.065 + (4 \times 15.999) = 24.305 + 32.065 + 63.996 = 120.366 \, \text{g/mol} $$

Finally, we calculate the mass of MgSO4:

$$ \text{mass} = 0.025 \, \text{moles} \times 120.366 \, \text{g/mol} = 3.00915 \, \text{grams} $$

So, approximately 3.01 grams of MgSO4 are needed to prepare 125 mL of 0.200 M magnesium sulfate solution.