You have a solution that is 12.0% sucrose (C₁₂H₂₂O₁₁) by weight (w/w) and you want 35.1 mmol of sucrose.

How many grams of this solution do you need?

Question

You have a solution that is 12.0% sucrose (C₁₂H₂₂O₁₁) by weight (w/w) and you want 35.1 mmol of sucrose.

How many grams of this solution do you need?

| General Chemistry 2

Accepted Answer

n_sucrose = $\frac{m_{sucrose}}{M_{sucrose}}$ with M_sucrose = 12 M_C + 22 M_H + 11 M_O = 12 x 12.0 + 22 x 1.01 + 11 x 16.0 = 342 g.mol^-1

12.0% sucrose by weight (w/w): m_sucrose = $\frac{12}{100}$ x m_solution

Mass of this solution needed:

m_solution = $\frac{100}{12}$ x m_sucrose

m_solution = $\frac{100}{12}$ x n_sucrose x M_sucrose

m_solution = $\frac{100}{12}$ x 35.1 x 10^-3 x 342

m_solution = 1.00 x 10² g

Midterm 1 - Question 3 | General Chemistry 2