## Introductory Chemistry (5th Edition)

Mass percent = (Mass solute X 100)/ (Mass solute + Mass solvent) (a) 41.2 g $C_{12}H_{22}O_{11}$ in 498 g $H_{2}O$ = (41.2 X 100)/ (498 + 41.2) = 4120/539.2 = 7.640 % (b) 178 mg $C_{6}H_{11}O_{6}$ in 4.91 g $H_{2}O$ = (178 X 10^-3 X 100)/ (0.178 + 4.91) = 17.8 / 5.088 = 3.498 % (c) 7.55 g NaCl in 155 g $H_{2}O$ = (7.55 X 100)/(7.55+155) = 755/162.55 = 4.64 %