# Calculate the molarity and the molality of an NH3 solution made up of 30.0 grams of NH3 in 70.0...

## Question:

Calculate the molarity and the molality of an {eq}NH_3 {/eq} solution made up of 30.0 grams of {eq}NH_3 {/eq} in 70.0 grams of water. The density of the solution is 0.982 g/mL.

## Molarity and Molality

Molarity and molality are ways to express the concentration of a solution. Molarity is the ratio between the moles of solute and the volume of the solution in liters. Molality is the ratio between the moles of solute and the mass of solvent in kilograms.

{eq}\rm Molarity \, (M) \, = \, \dfrac{Moles \ solute}{Volume \ of \ solution \ (L) } \\[4ex] \rm Molality \, (m) \, = \, \dfrac{Moles \ solute}{Mass\ of \ solvent \ (kg)} {/eq}

The symbol for molarity is {eq}\rm M {/eq} or {eq}\rm c {/eq} with the units also as {eq}\rm M {/eq} or {eq}\rm mol/L {/eq}.

The symbol for molality is {eq}\rm m {/eq} or {eq}\rm b {/eq} with the units also as {eq}\rm m {/eq} or {eq}\rm mol/kg {/eq}.

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We are given:

• Mass of solute ({eq}\rm NH_3 {/eq}) = 30.0 g.
• Mass of solvent = 70.0 g.
• {eq}\rm \rho {/eq} solution = 0.982 g/mL.

First, we will...