Question

What is the molarity of a 511 ml solution containing
205 mg of NH3?

Answer 1

Answer: The molarity of given solution is 0.0235 M.

Step-by-step explanation:

Given : Volume of solution = 511 mL

Convert mL into L as follows.

`1 mL = 0.001 L\\511 mL = 511 mL * (0.001 L)/(1 mL)\\= 0.511 L`

Mass of
`NH_(3)` (solute) = 205 mg

Convert mg into gram as follows.

`1 mg = 0.001 g\\205 mg = 205 mg * (0.001 g)/(1 mg)\\= 0.205 g`

As molar mass of
`NH_(3)` is 17 g/mol. Hence, number of moles of
`NH_(3)` are calculated as follows.

`No. of moles = (mass given)/(molar mass)\\= (0.205 g)/(17 g/mol)\\= 0.012 mol`

Molarity is the number of moles of a solute dissolved in a liter of solution.

`Molarity = (no. of moles)/(Volume (in L))`

Substitute the values into above formula as follows.

`Molarity = (no. of moles)/(Volume (in L))\\= (0.012 mol)/(0.511 L)\\= 0.0235 M`

Thus, we can conclude that the molarity of given solution is 0.0235 M.