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

Answer:

m = 0.0035 m.

Explanation:

Hello there!

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

$m=\frac{n_{solute}}{m_{solvent}}$

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

$n=30mgK_3PO_4*\frac{1gK_3PO_4}{1000gK_3PO_4}*\frac{1molK_3PO_4}{212.27gK_3PO_4} =1.41x10^{-4}mol$

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):

$m=\frac{1.41x10^{-4}mol}{0.0400kg}\\\\m=0.0035m$

In molal units (m=mol/kg).

Best regards!