Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.75 L of a 0.130 M sucrose solution is allowed to react for 190 minutes.

Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.

Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.

From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.

So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-24T09:08:07+01:00

The mass of sucrose that is obtained after 190 minutes is 106 g.

The following information in the complete question;

Rate constant (k) = 1.8 x 10^-4 s^-2

Initial concentration (Ao) = 0.130 M

Time taken = 190 minutes or 11400 seconds

Final concentration (A) = ?

Given that;

lnA = lnAo - kt

A = e^ lnAo - kt

A = e^[(ln0.130) - ( 1.8 x 10^-4 x 11400)

A = 0.017 M

Now;

n = CV

n = Number of moles

C = Molar concentration

V = volume

number of moles after hydrolysing = 0.017 M × 2.75 L

= 0.047 moles

Number of moles before hydrolysing =0.130 M × 2.75 L = 0.358 moles

Number of moles hydrolysed = 0.358 moles - 0.047 moles = 0.311 moles

Mass = Number of moles × molar mass

Molar mass of sucrose = 342 g/mol

Mass = 0.311 moles × 342 g/mol

Mass = 106 g

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