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.
I dont know but do you know da wae brudda?
Answer: 178.9 g
Explanation:
Density = 
find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³
1.0 cm³ = 1.0 mL
so 195.112 cm³ = 195.112 mL
plug value into density equation:
0.917 g/mL = (mass) / (195.112 mL)
and solve for mass!
The concentration of the solution is 4.25 M
Explanation
molarity=moles/volume in liters
moles = mass/molar mass
molar mass of HF = 19 + 1 = 20 g/mol
moles is therefore = 17.0 g/ 20 g/mol = 0.85 moles
volume in liters = 2 x10^2ml/1000 = 0.2 liters
therefore molarity = 0.85/0.2 = 4.25 M
