%yield = 91.8
<h3>Further explanation</h3>
Given
20 g NaCl
45 g AgCl
Required
%yield
Solution
Reaction
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
mol NaCl :
= mass : MW
= 20 g : 58,44 g/mol
= 0.342
mol AgCl from equation :
= 1/1 x mol NaCl
= 1/1 x 0.342
= 0.342
Mass AgCl(theoretical) :
= mol x MW
= 0.342 x 143,32 g/mol
= 49.02 g
%yield = (actual/theoretical) x 100%
%yield = (45/49.02) x 100%
%yield = 91.8
I think it would be false hope this helps
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer
pH=8.5414
Procedure
The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.
pH = pKa + log₁₀ ([A⁻] / [HA])
Where
pH = acidity of a buffer solution
pKa = negative logarithm of Ka
Ka =acid disassociation constant
[HA]= concentration of an acid
[A⁻]= concentration of conjugate base
First, calculate the pKa
pKa=-log₁₀(Ka)= 8.6383
Then use the equation to get the pH (in this case the acid is HBrO)
