How many grams of sucrose (C12H22O11) are in 1.55 L of 0.758 M sucrose solution?

Mass of Sucrose in given Solution is 401.8158 grams.

Explanation:

Given,

Molarity  of Sucrose Solution = 0.758 M.

Volume of Sucrose Solution(V) = 1.55 L.

To find the mass of Sucrose in the solution, we need to find the number of moles of Sucrose present in the given solution 0f 0.758 M.

c = 0.758 M

number of moles can be calculated using

n = c*V

n= 0.758*1.55  = 1.1749 moles

Then to convert no, of moles to mass, we can use the formula,

number of moles(n) = Mass(m)/Molar Mass(M)

The Molar mass of Sucrose C₁₂H₂₂O₁₁ is

Mass of Carbon Sucrose = 12*12 = 144  grams

Mass of Hydogen in Sucrose = 1*22 = 22 grams

Mass of Oxygen in Sucrose = 16*11 = 176 grams

Molar mass of Sucrose = 144 + 22 +176 = 342 grams.

Substituting the values in the formula,

number of moles(n) = Mass(m)/Molar Mass(M)

we get,

1.1749 = (m)/342 => m = 1.1749 *342 = 401.8158 grams.