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Katena32 [7]
2 years ago
9

Calculate the percentage of yield when 20 grams of sodium chloride solution reacts with an excess amount of silver nitrate solut

ion knowing that 45 grams of silver chloride precipitated​
Chemistry
1 answer:
Lady bird [3.3K]2 years ago
7 0

%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

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Answer : The molar concentration of sucrose in the tea is, 0.0549 M

Explanation : Given,

Mass of sucrose = 3.765 g

Volume of solution = 0.200 L

Molar mass of sucrose  = 342.3 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of sucrose}}{\text{Molar mass of sucrose}\times \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.765g}{342.3g/mole\times 0.200L}=0.0549mole/L=0.0549M

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