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3 years ago

A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are

Chemistry

1 answer:

Butoxors [25]3 years ago

7 0

<h3>Answer:</h3>

82.11%

<h3>Explanation:</h3>

We are given;

Theoretical mass of the product is 137.5 g
Actual mass of the product is 112.9 g

We are supposed to calculate the percentage yield

We need to know how percentage yield is calculated;
To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.

Thus;

% yield = (Actual mass ÷ Experimental mass) × 100%

= (112.9 g ÷ 137.5 g) × 100%

= 82.11%

Therefore, the percentage yield of the product is 82.11 %

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$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of estrogen = 13.5 g

Molar mass of estrogen = 272.40 g/mol

Volume of solution = 181 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{13.5\times 1000}{272.40\times 181}\\\\\text{Molarity of solution}=0.274M$

Hence, the molarity of solution is 0.274 M

To calculate the osmotic pressure of the solution, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = 0.274 M

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$\pi=1\times 0.274mol/L\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\pi=6.70atm$

Hence, the osmotic pressure of the solution is 6.70 atm

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