1.

Let a fully charged lead-storage battery contains 1.5 L of 5MH_2SO_2. What will be the concentratisof H_2SO_4 in the battery after 2.5 ampere current is drawn from the battery for 6 hour?

Answer»

4.626M
`0.1865 M `
`0.373 M`
`9.627 M `

Solution :`1F to 1 ` MOLE of `H_2SO_4 ` , 96500 CaCl ` to ` 1 mole
` 2.5 xx 6 xx 60 xx 60 to n = ((2.5 xx 6 xx 60 xx 60 xx 1)/(96500)) = 0.5596` moles
` n = (1.5 xx 5-n) =6.9404 , M^(1) = (n^(1))/(V) = (6.9404)/(1.5) = 4.6269M`


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