Answer:
Mole fraction: 0,0157.
Molality: 0,889m
Mass%: 16%
Explanation:
The units are mole fraction, molality and mass%
Units of mole fraction are moles of glucose per total moles.
Moles of glucose assuming 1L are:
0,944 moles.
Moles of water in 1L are:
1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = 59,0 moles of water
Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = 0,0157
Molality is mole of solute (0,944) per kg of solution (1,0624kg):
0,944mol / 1,0624kg = 0,889m
In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:
0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:
170g / 1062,4g ×100 = 16%
I hope it helps!
The mole fraction be "0.0187", molality be "1.06 mol/kg" and mass be "0.16".
Density = 1.0624 g/mL
Glucose = 0.944 M
Temperature = 20°C
fw of C₆H₁₂O₆ = 180.2 g/mol
As we know,
→ mass of A = (moles of A) x (fw of A)
= (Molarity of A) x V x (fw of A)
By substituting the values,
= [tex]0.944\times V\times 180.2[/tex]
= [tex]170.1 \ V g[/tex]
Now,
mass of (A + B) = density of solution x V
= [tex]1062.4 \ g/L\times V[/tex]
= [tex]1062.4\ Vg[/tex]
(1) [tex]Mole \ fraction = \frac{(Molarity \ of\ A\times V)}{[(Molarity \ of \ A\times V)+ \frac{(Mass \ of (A + B) - Mass \ of \ A)}{18} ]}[/tex]
By substituting the values, we get
[tex]= \frac{0.944V}{0.944V +\frac{(1062.4V-170.1V)}{18} }[/tex]
[tex]= 0.0187[/tex]
(2) The molality be:
= [tex]\frac{Moles \ of \ A}{Kg \ of \ B}[/tex]
= [tex]\frac{(0.944\times V)}{\frac{(1062.4V-170.1V)}{1000} }[/tex]
= [tex]1.06 \ mol/kg[/tex]
(3) The mass percentage be:
= [tex]\frac{Mass \ of \ A}{Mass \ of (A + B)}[/tex]
= [tex]\frac{170.1}{1062.4}[/tex]
= [tex]0.16[/tex]
Thus the above responses are correct.
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