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kupik [55]
3 years ago
10

What mass of sucrose (C12H22O11) should be combined with 488 g of water to make a solution with an osmotic pressure of 8.00 atm

at 290 K? (Assume the density of the solution to be equal to the density of the solvent.)
Chemistry
2 answers:
Soloha48 [4]3 years ago
7 0

Answer:

We need to add 51.52 grams of sucrose

Explanation:

Step 1: Data given

Mass of water = 488 grams

Density = 1g/mL

osmotic pressure = 8.00 atm

Temperature = 290 K

Step 2: Calculate concentration

π = i*M*R*T

8.00 atm = 1* M*0.08206 * 290

M = 0.336 M

Step 3: Calculate volume of water

488 grams = 0.488 L

Step 4: Calculate moles sucrose

moles sucrose = molarity * volume

Moles sucrose = 0.336 M * 0.448 L

Moles sucrose = 0.1505 moles

Step 5: Calculate mass sucrose

Mass sucrose = 0.1505 moles * 342.3 g/mol

Mass sucrose = 51.52 grams

We need to add 51.52 grams of sucrose

VikaD [51]3 years ago
5 0

Answer:

56 g of sucrose is the mass needed

Explanation:

Formula for osmotic pressure → π = M . R . T

8 atm = M . 0.082 L.atm/mol.K . 290 K

8 atm / (0.082 L.atm/mol.K . 290 K) = M → 0.336 mol/L

Let's determine the mass of sucrose that represents 0.336 mol

0.336 mol . 342 g / 1mol = 114.9 g

This is the mass that corresponds to 1L of solution, but we have 0.488 L

Solution density = 1 g/mL → 488 g are contained in 488 mL.

488 mL . 1L / 1000 mL = 0.488 L

Let's make a rule of three:

1L is the volume for 114.9 g of sucrose

In 0.488 L of volume, we need a mass of (0.488L . 114.9 g) 1L = 56 g

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

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