<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m
<u>Explanation:</u>
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute (KOH) = 3.301 g
= Molar mass of solute (KOH) = 56.1 g/mol
= Mass of solvent = 96.699 g
Putting values in above equation, we get:
Hence, the molality of potassium hydroxide solution is 0.608 m
Answer:
37400000000 is the <em>ans</em>
Percent composition by mass is calculated (mass of element within compound)/(mass of compound)*100. The lower the total molar mass of the compound, the greater the percent composition of sulfur. In this case, MgS would be that compound, since Mg has the lowest molar mass of the four elements bonded to S.
Answer:
Hydrogen = 2.5 * 10^21
Explanation:
Chemical Formula Glucose: C₆H₁₂O₆
One of the ways you could do this is to notice that for every carbon atom there are two Hydrogen atoms. You can state this more formally by using the formula to set up a ratio: 12/6 = hydrogen to Carbon
So if there are 1.250 * 10^21 Carbon atoms in the Glucose sample, then there will be twice as many hydrogen atoms.
H = 2 * 1.25 * 10^21 = 2.5 * 10^21 atoms
You could do this more formally by setting up a proportion.
6 Carbon / 12 Hydrogen = 1.25*10^21 / x Cross Multiply
6*x = 12 * 1.25*10^21 Combine the right
6x = 1.5 * 10^22 Divide by 6
x = 2.5 * 10^21
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
