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Sagot :
Answer: a) percent by mass = [tex]70.5\%[/tex]
b) molality = 38.01mol/kg
c) mole fraction = 0.40
Explanation:
Given molarity = 15.9 M
moles of [tex]HNO_3[/tex] = 15.9 moles in 1.00 L of solution
mass of [tex]HNO_3[/tex] = [tex]moles\times {\text {Molar mass}}=15.9mol\times 63g/mol=1001.7g[/tex]
Density of solution of = 1.42 g/ml
Volume of solution = 1.00 L = 1000 ml
Mass of solution = [tex]Density\times volume= 1.42g/ml\times 1000ml=1420g[/tex]
mass of solvent = mass of solution - mass of solute = (1420-1001.7) g = 418.3 g = 0.4183 kg
moles of solvent = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{418.3g}{18g/mol}=23.2mol[/tex]
percent by mass =[tex]\frac{\text {mass of nitric acid}}{\text {mass of solution}}\times 100=\frac{1001.2}{1420}\times 100=70.5\%[/tex]
molality = [tex]\frac{\text {no of moles of solute}}{\text {mass of solvent in kg}}=\frac{15.9mol}{0.4183kg}=38.01mol/kg[/tex]
mole fraction = [tex]\frac{\text {moles of solute}}{\text {total moles}}=\frac{15.9}{15.9+23.2}=0.40[/tex]
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