Respuesta :
Answer:
density is 1.057 g/mL
mole fraction is 0.0180
molarity is 0.98 mol/L
mole fraction is 0.0180
Explanation:
Given data
acid mass = 10 g
water volume = 100 mL
total volume = 104 mL
density = 1 g/cm³
to find out
the density, mole fraction, molarity, and molality
solution
first we calculate the density that is = total mass g / volume of solution mL
total mass = mass of H3PO4 + water mass
so water mass = density x volume
water mass = 100 ml x 1.0 g/cm3
water mass = 100 g
so total mass = 110.00 g
so that
density = total mass g / volume of solution mL
density = 110 / 104 = 1.057 g/mL
now we calculate no of moles in solvent i.e = mass H2O / mlar mass H2O
no of moles in solvent = 100/18.015 = 5.55 moles
and no of moles in solute i.e = mass of H3PO4 / mlar mass in H3PO4
moles in solute i.e = 10/ 97.994 = 0.102 moles
so total moles is 5.55 + 0.102 = 5.652 moes
so now mole fraction = no of moles in solute / total moes
mole fraction = 5.55 / 5.65
mole fraction is 0.0180
now we calculate first
mole fraction in solute and solvent that is
mole fraction in solute = no of moles in solute/ total moles
mole fraction in solute = 0.102 /5.65
mole fraction in solute is 0.0180
and mole fraction in solvent that = no of moles in solvent/ total moles
mole fraction in solvent that = 5.55/ 5.65
mole fraction in solvent that is 0.982
so molarity = no of moles of solute / volume
molarity = 0.102/0.104
molarity is 0.98 mol/L
and molality is = no of moles of solute / mass
molality = 0.102 / 100
molality is 1.02 mol/kg
The density of the solution is defined as the mass per unit volume. The density, mole fraction, molarity, and molality of the solution are 1.057 g/mL,0.0180,0.98 mol/L,1.02 mol/kg respectively.
What is density?
Density is defined as the mass per unit volume. Its unit is kg/m³.it is used to find how the material is dense.
The given data in the problem are ;
Mass of acid= 10 g
volume of water = 100 mL
total volume given = 104 mL
Density of water = 1 g/cm³
(a) Density
[tex]\rm{ Density \;of \;solution=\frac{Total \;mass }{Total \;volume} }[/tex]
Total mass = mass of H₃PO₄ + mass of water
Mass of water = density of water × volume of water
Mass of water([tex]\rm m_w[/tex])= 1 g/cm³×100 mL
Mass of water([tex]\rm m_w[/tex])=100 g.
Total mass = mass of H₃PO₄ + mass of water
Total mass= 10g+100g= 110 g
[tex]\rm{ Density \;of \;solution=\frac{Total \;mass }{Total \;volume} }[/tex]
[tex]\rm{ Density \;of \;solution=\frac{110}{104} }\;gm/mL[/tex]
[tex]\rm{ Density \;of \;solution=1.057 gm/ml[/tex]
Hence the density of the solution will be 1.057 gm/mL.
(b)Mole fraction
Mole of solution = Mole of solvent + Mole of solute
[tex]\rm{ Mass \;of\; solvent = \frac{Mass\;of \;H_2O }{Moler \;Mass\; of\;H_2O} }\\\\\rm{ Mass \;of\; solvent = \frac{100 }{18.015}\;mole\\\\[/tex]
[tex]\rm{ Mass \;of\; solvent = 5.55 \;mole[/tex]
[tex]\rm{ Mass \;of\; solute = \frac{Mass\;of \;H_3PO_4 }{Moler \;Mass\; of \;H_3PO_4 } }\\\\\rm{ Mass \;of\; solute = \frac{10 }{97.9944}\;mole\\\\[/tex]
[tex]\rm{ Mass \;of\; solvent =0.102 \; mole[/tex]
Moles of solution = Mole of solvent + Mole of solute
Moles of solution = 5.55+0.102= 5.652 moles.
[tex]\rm Mole \; fraction= \frac{Mole\; of \;solute }{Mole \;of \;solution} \\\\\rm Mole \; fraction= \frac{5.55 }{5.65}}\\\\\rm Mole \; fraction= 0.0180[/tex]
Hence the mole fraction of the solution is 0.0180
(c) Molarity
[tex]\rm{Molarity= \frac{no \;of\; moles \;of \;solute}{volume} }\\\\\rm{Molarity= \frac{0.102}{0.104} }\\\\\rm{Molarity= 0.98 mol/lit}[/tex]
Hence the molarity of the solution will be 0.98 mol/lit.
(d) Molality
[tex]\rm Molality = \frac{No \;of \;moles \;of \;solute}{mass} \\\\\rm{Molality= \frac{0.102}{100} } \\\\\rm{Molarity=1.02 mol/kg }[/tex]
Hence the molality of the solution will be 1.02 mol/kg.
Hence the density, mole fraction, molarity, and molality of the solution will be is 1.057 g/mL,0.0180,0.98 mol/L,1.02 mol/kg respectively.
To learn more about the density refer to the link
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