1. Molar mass of CO(NO_{3})_{2} . 6H_{2}O = 58.7 + 2(14 + 48) + (6 x 8)g mol^{-1}

= 58.7 + 124 + 108 g mol^{-1} = 290.7 g mol^{-1}

^{}

Volume of solution = 4.3 L

2. 1000 mL of 0.5 M H_{2}SO_{4} Contain H_{2}SO_{4} = 0.5 moles

30 mL of 0.5 M H_{2}SO_{4} contain H_{2}SO_{4} = \(\frac{0.5}{100}\) x 30 mole

= 0.015 mole

Volume of solution = 500 mL = 0.500 L