## General Chemistry: Principles and Modern Applications (10th Edition)

(a) 1. Calculate the molar mass: $C_{12}H_{22}O_{11}$ : ( 1.008 $\times$ 22 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 11 )= 342.30 g/mol 2. Use the informations as conversion factors to find the molarity of this solution: $$\frac{ 10.00 g \space C_{12}H_{22}O_{11} }{100 \space g \space solution} \times \frac{1 \space mol \space C_{12}H_{22}O_{11} }{ 342.30 \space g \space C_{12}H_{22}O_{11} } \times \frac{ 1.040 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.3038 \space M$$ (b) By definition: 10.00 g of sucrose = 100 g of solution mass of water = 100 - 10.00 = 90.00 g of water = 90.00 $\times 10^{-3}$ kg of water. moles of sucrose: $$10.00 \space g \times \frac{1 \space mol}{342.30 \space g} = 0.02921 \space mol$$ $$Molality = \frac{0.02921 \space mol \space sucrose}{90.00 \times 10^{-3} \space kg \space water} = 3.246 \space m \space sucrose$$ (c) Using the same definition as in (b): $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol moles of water = $90.00 \space g \times \frac{1 \space mol}{18.02 \space g} = 4.994 \space mol \space water$ $$x_{sucrose} = \frac{0.02921}{4.994 + 0.02921} = 0.005815$$