Molarity Practice

Question

6. Complete the table:

___________________________________________________________________

| Solute | Mass of solute | Moles of solute | Volume of solution | Molarity of solution |

| MgSO₄ | 0.538 g | ____________ | 29.0 mL | _________________|

| NaOH | ____________ | ____________ | 105.0 mL | 1.45 M |

| CH₃OH | 14.0 g | _____________ | _____________ | 0.520 M |

Complete the row of MgSO₄.

Enter your answers numerically separated by a comma.

Mass of solute, molarity of solution = ____________________ mol, M

Answer 1

MgSO₄ row: 4.47 × 10⁻³ mol, 0.154 M

Moles of solute

M(MgSO₄) = 24.31 + 32.06 + 4(16.00) = 120.4 g/mol

n = 0.538 g / 120.4 g/mol = 4.47 × 10⁻³ mol

Molarity

Convert the volume to litres first: 29.0 mL = 0.0290 L

M = 4.47 × 10⁻³ mol / 0.0290 L = 0.154 M

A note on the prompt

The answer box says "Mass of solute, molarity of solution," but the mass (0.538 g) was already given to you and the blanks in that row are moles and molarity. That looks like a labelling slip in the assignment — enter 4.47 × 10⁻³, 0.154, matching the "mol, M" units printed after the blank.

The rest of the table, since it is all the same two steps

NaOH row (105.0 mL, 1.45 M):
n = (0.1050 L)(1.45 M) = 0.152 mol
M(NaOH) = 22.99 + 16.00 + 1.01 = 40.00 g/mol
m = (0.152)(40.00) = 6.09 g

CH₃OH row (14.0 g, 0.520 M):
M(CH₃OH) = 12.01 + 4(1.01) + 16.00 = 32.04 g/mol
n = 14.0 / 32.04 = 0.437 mol
V = 0.437 / 0.520 = 0.840 L = 840 mL

The pattern behind the whole table

Every cell is one of three rearrangements of the same two relationships:

n = m / M(molar mass)  and  M(molarity) = n / V

Moles sit in the middle of both, which is why the moles column is the hinge — whatever you are given, get to moles first and the remaining blank falls out in one step.

The error to watch: volume must be in litres every time. Using 29.0 instead of 0.0290 would give 1.54 × 10⁻⁴ M, off by a thousand. Convert as you write the number down rather than at the end.