We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Answer:
Order, sensitivity or response to the environment, reproduction, growth and development, regulation, homeostasis, and energy processing.
Answer:
She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.
Explanation:
A supersaturated solution contains more salt than it can normally hold at a given temperature.
A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.
Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.
B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.
C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.
Answer:
10grams
Explanation:
So it weighs 236 grams added with 25 grams. So it now weighs 261 grams so 10 grams of sugar remains in it.
Answer:
.615
Explanation:
Please refer to the attached file for the detailed step by step solution to the given problem.
