Calculate the molarity of each of the following solutions:

Explanation

(a) 30 g of `Co (NO_(3)) _(2). 6H_(2) O ` in `4.3L` of solution
(b) 30 mL of `0.5 M H _(2) SO _(4)` diluted to 500 mL.
(a) Molar mass of `Co(NO _(3))_(2). 6H_(2)O = 58.7 + (14 + 48) + (6 xx 18) g mol ^(-1)`
`= 58.7 + 124 + 108 mol ^(-1) = 290.7 g mol ^(-1)`
No. of moles of `Co(NO_(3)) _(2). 6H_(2)O = ("Mass")/("Molar mass") = (30g)/(290. 7 g mol ^(-1)) = 0.103`
Volume of solution `= 4.3 L`
`therefore` Molarity of solution `= ("No.of moles of solute")/("Volume of solution in L")`
`= (0.103 "mole")/(4.31)= 0.024 M`
(b) 1000 mL of `0.5 M H_(2) SO_(4)` Contain `H_(2) SO_(4) =0.5` moles
`therefore 30 mL` of `0.5 M H _(2) SO _(4)` contain `H_(2) SO _(4)= (0.5)/(1000)_ xx 30` mole `00.015 ` mole
Volume of solution `= 500 mL = 0.500 L`
`therefore` Molarity of solution `= ("No. of moles of solute")/("Volume of solution in L") = (0.015)/(0.5) =0.03 M`