# Calculate (he molarity of each of the following solutions: 1. 30 g of CO(NO3)2 . 6H2O = in 4.3 L of solution

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Calculate (he molarity of each of the following solutions:

1. 30 g of CO(NO3)2 . 6H2O = in 4.3 L of solution

2. 30 mL of 0.5 M H2SO4 diluted to 500 mL.

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1. Molar mass of CO(NO3)2 . 6H2O = 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 H2SO4 Contain H2SO4 = 0.5 moles

30 mL of 0.5 M H2SO4 contain H2SO4$\frac{0.5}{100}$ x 30 mole

= 0.015 mole

Volume of solution = 500 mL = 0.500 L