Question

What is solute mass and mol solute of `NaHCO_3` which has the volume solution 250.0 mL and molarity 0.100 mol/L?

Answer 1

`a. eta=0.025mol`
`b. m=2.10g`

Explanation:

Given the molarity `(M)` and the volume `(V)` of the solution, the number of moles solute`(eta)` can be calculated through the formula shown below. Rearrange the formula to isolate the unknown variable, the `eta`;

`M=eta/V`
`eta=MxxV`

where:

`M=(0.100mol)/(L)`; the concentration
`V=250.0ml`

`eta=(0.100mol)/cancel((L))xx250.0cancel(ml)xx(1cancel(L))/(1000cancel(ml))`
`eta=0.025mol`

Now, since `eta` is already obtained, mass of the solute can also be calculated using the formula described below. Rearrange formula to isolate the unknown variable.

`eta=("mass"(m))/("Molar mass"(Mm))`
`m=etaxxMm`

where:

`eta=0.025mol`; as obtained from the preceding calculation
`Mm=(84.01g)/(mol)`; is obtainable from the periodic table.

`m=0.025cancel(mol)xx(84.01g)/cancel(mol)`
`m=2.10g`