<u>Answer:</u> The molality of solution is 0.740 m.
<u>Explanation:</u>
To calculate the mass of solvent (water), we use the equation:

Volume of water = 750 mL
Density of water = 1 g/mL
Putting values in above equation, we get:

To calculate the molality of solution, we use the equation:

Where,
= Given mass of solute
= 100.0 g
= Molar mass of solute
= 180 g/mol
= Mass of solvent (water) = 750 g
Putting values in above equation, we get:

Hence, the molality of solution is 0.740 m.
Water
Water is a compound because it is made from more than one kind of element (oxygen and hydrogen).
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
1 mol = 6.023x10^23 number of molecules (Avogadro's number)
1 : 6.023x10^23
X : 4.91x10^22
(6.023x10^23)X = 4.91x10^22
X = 4.91x10^22/6.023x10^23
X = 0.082 Moles
Answer:
Examples of radioactive isotopes include carbon-14, tritium (hydrogen-3), chlorine-36, uranium-235, and uranium-238.