Such an object makes a larger dent in the fabric of space-time than an object with little mass. (It has a greater gravitational attraction than less massive objects)
A greater force is required to accelerate such an object than a less massive object
Answer:
1367.7 g of ethylene glycol was added to the solution
Explanation:
In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')
525.8 mmHg - 451 mmHg = 451 mmHg . Xm
74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)
Xm = Mole fraction of solute / Moles of solute + Moles of solvent
We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol
(Notice we converted the 2kg of water to g)
0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent
0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute
18.4 moles = Moles of solute - 0.166 moles of solute
18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles
Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g
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Answer:
520ML and apparently I need to put more in this answer
Explanation:
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