If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln
2 answers:
Answer:
Explanation:
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Molality is defined as:
Thus, as the solution is % by mass, one assumes that the mass of urea is 26g and the mass of the solution 100g, in such a way, one computes the moles of urea by knowing its molecular mass:
Now, as solution has a mass of 100g, the solvent has a mass of 74g which corresponds to 0.074L since the density of water is 1g/mL; in such a way the molality is:
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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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Hello! The correct answer is 1. KCI. I really hope this helped you out! c:
<span>The </span>equilibrium<span> will </span>shift<span> to favor the side of the reaction that involves fewer moles of gas. Its C </span>
0,5 kilometers = 500 metres 500<600 Maria ran farther
means 0.5 raise to the power of n. "To the power of" means an action of multiplication of 0.5 to n times.
The different values of n are given in the table. Substituting these values:
For A:
n = 1 (given)
For B:
n = 2 (given)
For C:
n = 3 (given)
For D:
n = 6 (given)
For E:
n = 8 (given)
Hence, the values are:
A = 0.5
B = 0.25
C = 0.125
D = 0.015625
E = 0.00390625
10-7.5=2.5, 2.5/2=1.25, 7.5+1.25=8.75