Đáp án:
`C% = {m_(ct)}/{m_(dd)}`
`⇒ m_(ct) = C% . m_{dd}`
`m_{H_2SO_4} = 500 . 9,5% = 47,5` `gam`
`n_{H_2SO_4} = (47,5)/(98) = (95)/(196) = 0,48` `mol`
`m_{H_2SO_4} = \frac{500 xx 9,5}{100} = 47,5` `(g)`
`->` Số mol của `H_2SO_4`: `\frac{47,5}{98} \approx 0,484` `(mol)`
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