Question

How many grams of sucrose (c12h22o11) are in 1.55 l of 0.758 m sucrose solution? express your answer in grams to three significant figures?

Answer 1

Answer : The mass of sucrose are, 402 grams

Solution :

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

$M=\frac{w}{M\times V}$

where,

M = molarity of solution = 0.758 M = 0.758 mole/L

w = mass of solute (sucrose) = ?

M = molar mass of solute (sucrose) = 342 g/mole

V = volume of solution in liter = 1.55 L

Now put all the given values in the above formula, we get the mass of sucrose.

$0.758mole/L=\frac{w}{342g/mole\times 1.55L}$

$w=402g$

Therefore, the mass of sucrose are, 402 grams

Answer 2

403.6 grams of sucrose (C12H22O11) are in 1.55L of 0.758M sucrose solution.

HOW TO CALCULATE MASS:

• The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

• However, the number of moles in the sucrose solution needs to be calculated as follows:

• no. of moles = molarity × volume

• no. of moles of sucrose = 0.758 × 1.55

• no. of moles = 1.18 moles

• Molar mass of C12H22O11 = 12(12) + 1(22) + 16(11)

• Molar mass of C12H22O11 = 144 + 22 + 176

• Molar mass of C12H22O11 = 342g/mol

• Mass of sucrose = 342g/mol × 1.18

• Mass of sucrose = 403.6g

• Therefore, 403.6 grams of sucrose (C12H22O11) are in 1.55L of 0.758M sucrose solution.

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