Calculate moles from provided weight

4. Moles and Molarity 4.1) Moles 4.1.1) Moles - General

4.1.1.1) Moles - General Normal 1


Total tried:       Correct:       Wrong:

A compounder weighed 113.68 mg of sodium lactate . If the molecular weight of sodium lactate is 112 g/mol, then determine the number of milimoles present there.


Click on the button below to see the answer and explanations

lb equals 1.015 mmol kg

    The molecular weight of sodium lactate is 112 g/mol, which means `(112 \quad g)/(1 \quad mol)`. Again, 113.68 mg equals `113.68/1000=0.11368` g of sodium lactate . Therefore, using equation method we can write:
    `(112 \quad g)/(1 \quad mol)=(0.11368 \quad g)/x therefore x = 0.001015` mol.
    Multiplying this number with 1000 will give us milimoles;
    `therefore 0.001015 × 1000 = 1.015` mmol. Ans.

    Using shortcut: simply divide the miligram weight (`113.68 \quad mg`) with molecular weight (`112\quad g/(mol)`). This will give us:
    `(113.68 \quad mg)/(112 \quad g/(mol))=1.015 \quad mmol` Ans.