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Yuliya22 [10]
3 years ago
9

What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm

at 290 K ? (Assume the density of the solution to be equal to the density of the solvent.)
Chemistry
1 answer:
lesya [120]3 years ago
5 0

<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 8.80 atm

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

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL    (Density of water = 1 g/mL)

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

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

Hence, the mass of sucrose required is 69.08 g

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In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of wate
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Answer:

21.344%

Explanation:

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The percent error = [(Actual -Experiment)/Actual]*100%

The percent error =  [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%

8 0
3 years ago
ithium metal reacts with water to give lithium hydroxide and hydrogen gas. if 75.5 mL of hydrogen gas is produce at STP, what is
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The reaction equation may be written as:
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6 0
3 years ago
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2. Determine the heat of reaction (AH,xn) for the process by which hydrazine (N2H4)
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The heat of reaction : 50.6 kJ

<h3>Further explanation</h3>

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Reaction

N₂(g) + 2H₂(g) ⇒N₂H₄(l)

thermochemical data:

1. N₂H₄(l)+O₂(g)⇒N₂(g)+2H₂O(l)   ΔH=-622.2 kJ

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We arrange the position of the elements / compounds so that they correspond to the main reaction, and the enthalpy sign will also change

1. N₂(g)+H₂O(l) ⇒  N₂H₄(l)+O₂(g) ΔH=+622.2 kJ

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Add reaction 1 and reaction 2, and remove the same compound from different sides

1. N₂(g)+2H₂O(l) ⇒  N₂H₄(l)+O₂(g) ΔH=+622.2 kJ

2.2H₂(g)+O₂(g)⇒2H₂O(l)            ΔH=-571.6 kJ

-------------------------------------------------------------------- +

N₂(g) + 2H₂(g) ⇒N₂H₄(l)   ΔH=50.6 kJ

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