Question

How many moles of solute are there in the following2.90×10^-2 m Na2CrO4 solution made by dissolving the Na2CrO4 in 1000.0 g of water: _____ mol

Answer 1

2.90 x 10⁻² moles of solute (Na2CrO4).

Step-by-step explanation:

To solve this problem, we have to understand the concept of molality: The molality (m) of a solution is the moles of solute divided by the kilograms of solvent. The formula of molality is:

`Molality\text{ \lparen m\rparen=}\frac{moles\text{ of solute}}{kilograms\text{ of solvent}}=(mol)/(kg).`

As we have the molality 2.90 x 10⁻² m, but the mass in g and not kg, we have to convert this value. Remember that 1 kg equals 1000 g:

`1000.0\text{ g}\cdot\frac{1\text{ kg}}{1000\text{ g}}=1\text{ kg.}`

So we have 1 kg of solvent (water). Let's solve for 'moles of solute' and replace the data that we have, like this:

`\begin{gathered} Mole\text{s of solute=molality}\cdot kilograms\text{ of solvent,} \\ Moles\text{ of solute=2.90}\cdot10^(-2)\text{ m}\cdot1\text{ kg,} \\ Moles\text{ of solute=2.90}\cdot10^(-2)\text{ moles.} \end{gathered}`

The answer would be that we have 2.90 x 10⁻² moles of solute (Na2CrO4).