Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :

where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)

Therefore, the molal freezing point depression constant of X is 
Answer:
= konstanta produk kelarutan
= kation dalam larutan berair
= anion dalam larutan berair
= konsentrasi relatif a dan b
DARI WEB
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g . Solubility of NaNO3=21.9g or NaNO3 x 100 g/ 25 g =87.6. Calculate the molar mass of the dissolved compound as the sum of mass of all atoms in the molecule.
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Answer:
This question is incomplete.
Explanation:
This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;
Molarity = number of moles (n) / volume (in liter or dm³)
To calculate the number of moles of NaC₂H₃O₂, we say
number of moles (n) =
given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂
The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000
The unit for Molarity is M (Molar concentration), mol/L or mol/dm³