Answer:
m = 0.0035 m.
Explanation:
Hello there!
In this case, since the formula for the computation of the molality is:
[tex]m=\frac{n_{solute}}{m_{solvent}}[/tex]
We can first compute the moles of solute, K3PO4 by using its molar mass:
[tex]n=30mgK_3PO_4*\frac{1gK_3PO_4}{1000gK_3PO_4}*\frac{1molK_3PO_4}{212.27gK_3PO_4} =1.41x10^{-4}mol[/tex]
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):
[tex]m=\frac{1.41x10^{-4}mol}{0.0400kg}\\\\m=0.0035m[/tex]
In molal units (m=mol/kg).
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