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Calculate the molarity of each of the following solutions : (a) 30 g of Co(NO_3)_2. 6H_2O in 4.3 L of solution (b) 30 mL of 0.5 M H_2SO_4 diluted to 500 mL. |
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Answer» SOLUTION :(a) Molar mass of` Co(NO_3)_2.6H_2 O = 58.7 + 2(14 + 48) + 6 xx 18 = 310.7 g "mol"^(-1)` Number of moles of `Co(NO_3)_(2.6)H_2O = (30G)/(310 g "mol"^(-1) ) = 0.0966 `mol Volume of solution = 4.3 L MOLARITY of solution = 0.0966/4.3 =0.022 M. (b)1000 ML of 0.5 M `H_2SO_4` contain `H_2SO_4` = 0.5 mole ` therefore `30 mL of 0.5 M `H_2SO_4` contain `H_2SO_4`= 0.5/1000x30 mole = 0.015 mole Volume of solution = 500 mL = 0.500 L Molarity of solution =Number of moles of solute/Volume of solution in L `= (0.015)/(0.500) = 0.03 M` |
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