Question

A Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.50 LL of a 0.180 MM sucrose solution is allowed to react for 200 minmin .

Answer 1

The question is incomplete, here is the complete question:

The hydrolysis of sucrose into glucose and fructose in acidic water has a rate constant of 1.8×10⁻⁴ s⁻¹ at 25°C. Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.65 L of a 0.120 M sucrose solution is allowed to react for 200 min.

Answer: The mass of sucrose solution left is 12.56 grams

Step-by-step explanation:

Molarity is calculated by using the equation:

`\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}`

Molarity of sucrose solution = 0.120 M

Volume of solution = 2.65 L

`0.120mol/L=\frac{\text{Moles of sucrose solution}}{2.65L}\\\\\text{Moles of sucrose solution}=(0.120mol/L* 2.65L)=0.318mol`

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant =
`1.8* 10^(-4)s^(-1)`

t = time taken for decay process = 200 min = (200 × 60) = 12000 seconds

`[A_o]` = initial amount of the sample = 0.318 moles

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`1.8* 10^(-4)=(2.303)/(12000)\log(0.318)/([A])`

`[A]=0.0367mol`

To calculate the mass from given number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of sucrose = 0.0367 moles

Molar mass of sucrose = 342.3 g/mol

Putting values in above equation, we get:

`0.0367mol=\frac{\text{Mass of sucrose}}{342.3g/mol}\\\\\text{Mass of sucrose}=(0.0367mol* 342.3g/mol)=12.56g`

Hence, the mass of sucrose solution left is 12.56 grams