Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
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Answer:
Mole fraction of nitrogen = 0.52
Explanation:
Given data:
Temperature = 31.2 °C
Pressure = 870.2 mmHg
Volume = 15.1 L
Mass of mixture = 24.1 g
Mole fraction of nitrogen = ?
Solution:
Pressure conversion:
870.2 /760 = 1.12 atm
Temperature conversion:
31.2 + 273 = 304.2 K
Total number of moles:
PV = nRT
n = PV/RT
n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K
n = 16.9 L.atm. /25 L.atm. mol⁻¹
n = 0.676 mol
Number of moles of nitrogen are = x
Then the number of moles of CO₂ = 0.676 - x
Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)
24.1 = 28x + ( 29.7 -44x)
24.1 - 29.7 = 28x - 44x
-5.6 = -16 x
x = 0.35
Mole fraction of nitrogen:
Mole fraction of nitrogen = moles of nitrogen / total number of moles
Mole fraction of nitrogen = 0.35 mol / 0.676 mol
Mole fraction of nitrogen = 0.52
Answer:
3:1
Explanation:
Stearic acid - C18H36O2
Fructose- C6H12O6
Since the energy content of food is roughly proportional to the carbon content. It means:
C18/C6 = 3 or 3:1
This explains why fats and oil are the major source of energy in the body and not carbohydrates. This is due to Fats and oil having larger amounts of Carbon compared to Carbohydrates.
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
