Question

What is the molality of a solution that has 30mg of K3PO4 dissolved in 40mL of water? (The density of water is 1.00 g/mL)

Answer 1

m = 0.0035 m.

Step-by-step explanation:

Hello there!

In this case, since the formula for the computation of the molality is:

`m=(n_(solute))/(m_(solvent))`

We can first compute the moles of solute, K3PO4 by using its molar mass:

`n=30mgK_3PO_4*(1gK_3PO_4)/(1000gK_3PO_4)*(1molK_3PO_4)/(212.27gK_3PO_4) =1.41x10^(-4)mol`

Next, since the volume of water is 40.0 mL and its density is 1.00 g/mL we infer we have the same grams (40.0 g). Thus, we obtain the following molality by making sure we use the mass of water in kilograms (0.04000kg):

`m=(1.41x10^(-4)mol)/(0.0400kg)\\\\m=0.0035m`

In molal units (m=mol/kg).

Best regards!