Question

Calculate and explain (in words) how you would make 90.0 mL of 2.0 M MgSO4 solution from a solid solute MgSO4.

Answer 1

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

Answer 2

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:

$\displaystyle 90.0\text{ mL} \cdot \frac{2.0\text{ mol MgSO _4 }}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.18\text{ mol MgSO _4 }$

Convert this amount to grams. The molecular weight of MgSO₄ is 120.38 g/mol:

$\displaystyle 0.18\text{ mol MgSO _4 } \cdot \frac{120.38\text{ g MgSO _4 }}{1\text{ mol MgSO _4 }} = 22\text{ g MgSO _4 }$

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.