Answer to Question #347687 in General Chemistry for Amelia Kirlew

Question

Which of the following correctly determines the mole fraction of 250. g sucrose (MM = 342 g/mol) dissolved in 1000mL of water?

Answer 1

M_r(H₂O) = 18 g/mol

M_r(sucrose) = 342 g/mol

m(sucrose) = 250 g

p(H₂O) = 1 g/ml

mol fraction of sucrose = ?

m(H₂O) = p * V = 1g/ml * 1000 ml = 1000 g

n(H₂O) = m / M_r = 1000 / 18 = 55,56 mol

n(sucrose) = m / M_r = 250 / 342 = 0,731 mol

Mol fraction of sucrose = $\frac {n(sucrose)}{n(sucrose) + n(H_2O)} = \frac {0,731}{0,731+55,56} = 0,013$

ANSWER: 0,013 or 1,3%