Question

1. At 15 °C and 1atm, a series of Mg(NOs)2 aqueous solutions is prepared. It is found that the partial molar volume of Mg(NOs)2 (in mL/mol) fits the expression V(Mg(NOs)a2) = 70.21 b + 2.46 where b is the concentration of Mg(NO)h in molality (mol of solute/kg of solvent). (a) (10pt) Determine the partial molar volume of Mg(NO,)2 (in mL/mol) when 0.01800-mol Mg(NOs)2 mixes with 200.0 g of deionized water. dV(water) as a function of b from the partial molar volume expression db (b) (10pt) Derive the appropriate expression of of V(Mg(NO)2), where V(water) is the partial molar volume of water and b is the molality of Mg(NOs)2.

Answer 1

Molality of a solution is number of moles of solute dissolved in one Kg of solvent

molality

$m=\frac{n}{w}$

n is moles of solute

w is weight of solvent in Kg

Here at that VEMtogosh) 70.21(b) 2.46 melality Mg(No where b is a) Gairen M Nos 2 .018 mot e olvert 200g 0.016 Xi00a =) 0.09 00 VCrge) VMg(N 70.21 x 0.09 +2.46 8-7189 mt/mol that kmow fo We onstant temporatuore dve VA solution at A V> Volume Aelution VA- Pealial molar olar velume of A VR Potiat

number nA males A number o mols Vs ides dikrentiah beth On dv= dveydnt dn+ ) d ng dvs B Here d Vveten) db 0 db dvcwats) d70-21xbt2.46) 2 n, 2 woden db dviwaten) X 70.21 db dviwato) 70.2/X db

At given temperature and pressure partial molar volume is given by

$V=\frac{\partial V}{\partial n}$

where n is number of moles