Answer :
Answer:
See Below.
Explanation:
Recall that molarity is defined by moles of solute over liters of solution (mol/L).
Therefore, to make 90.0 mL of 2.0 M MgSO₄, we will need:
[tex]\displaystyle 90.0\text{ mL} \cdot \frac{2.0\text{ mol MgSO$_4$}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.18\text{ mol MgSO$_4$}[/tex]
Convert this amount to grams. The molecular weight of MgSO₄ is 120.38 g/mol:
[tex]\displaystyle 0.18\text{ mol MgSO$_4$} \cdot \frac{120.38\text{ g MgSO$_4$}}{1\text{ mol MgSO$_4$}} = 22\text{ g MgSO$_4$}[/tex]
Therefore, to make the solution, we can add 22 grams of MgSO₄ into a graduated cylinder, then mix and dilute the solution with distilled water until we reach 90.0 mL.
Answer:
See below
Explanation:
MgSO4 mole wt = 24.3+32+16*4 = 120.3 gm
for 90cc of 2M:
90/1000 * 2 * 120.3 = 21.66 gm
take 21.66 gm of the solid and dilute with water to 90 ml