Answer
Density = 1.06 g/cm$^3$
Mole fraction = 0.9820
Molarity = 0.981 mol/L
Molality = 1.02 mol/kg
Work Step by Step
First, calculate the density of H$_3$PO$_4$ using the density of water and its given mass at 100mL. Since the density of water is 1.00 g/mL the mass of 100 mL of water is equal to 100 grams.
Thus we may solve for the density of the solution,
$Density = \dfrac{mass}{volume} = \dfrac{10.0 \ g \ H_3PO_4 + 100.0\ g \ H_2O}{104 \ mL} = 1.06 \ g/mL = 1.06 g/cm^3$
Then we solve for the molar concentration of H$_3$PO$_4$ and H$_2$O
Mol H$_3$PO$_4$ $= 10.0 \ g \times \dfrac{1 \ mol}{ 97.99 \ g} = 0.102$ mol H$_3$PO$_4$
Mol H$_2$O $= 100. \ g \times \dfrac{1 \ mol}{ 18.02 \ g} = 5.55$ mol H$_2$O
We then solve for the mole fraction of H$_3$PO$_4$
Mole fraction of H$_3$PO$_4$ = $\dfrac{0.102 \ mol \ \text{H$_3$PO$_4$}}{(0.102 + 5.55) \ mol} = 0.0180$
Now that we know the mole fraction of H$_3$PO$_4$, we may solve for the mole fraction of H$_2$O,
Mole fraction of H$_2$O = 1.000 - 0.0180 = 0.920
Using the molar concentration, we may then solve for the molarity of H$_3$PO$_4$
Molarity $=\dfrac{0.102 \ mol H_3PO_4}{0.104 \ L} = 0.981 \ mol/L$
Then for the Molality of H$_3$PO$_4$,
Molality $=\dfrac{0.102 \ mol H_3PO_4}{0.100 \ kg} = 1.02 \ mol/kg$
