Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
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Its a 50% chance that approx 1/2 of the pennies will land on tails. The next toss will result the same. and so on and so on. showing how a reaction would slowly eliminate 1/2 of the remaining lives per reaction, until nothing is left. I think it is a good stimulation.
If I understand you correctly, you should fill the gap and decide which kind og bonds <span> allows metals to conduct electricity. If so, without any doubt I can say that the metalic bonds </span><span> allows metals to conduct electricity. I'm pretty sure it will help you! Regards.</span>
Answer:
1.12g/mol
Explanation:
The freezing point depression of a solvent for the addition of a solute follows the equation:
ΔT = Kf*m*i
<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>
<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>
<em>m is molality of the solution</em>
<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>
<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>
Replacing:
26.22°C = 5.35°Ckgmol⁻¹*m*1
4.90mol/kg = molality of the compound X
As the mass of the solvent is 100g = 0.100kg:
4.9mol/kg * 0.100kg = 0.490moles
There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:
0.551g / 0.490mol
= 1.12g/mol
<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>
