`m_{NaOH(1)} = \frac{50 xx 10}{100} = 5` `(g)`
`m_{NaOH(2)} = \frac{450 xx 25}{100} = 112,5` `(g)`
`a)`
`C%_{NaOH} = \frac{5 + 112,5}{50 + 450} xx 100% = 23,5%`
`b)`
$C_{M(NaOH)}$ `= \frac{10. 23,5. 1,05}{40} = 6,16875` `(M)`
`m_{NaOH} = \frac{(450 + 50). 23,5}{100} = 117,5` `(g)`
`->` `n_{NaOH} = \frac{117,5}{40} = 2,9375` `(mol)`
`->` $V_{dd(NaOH)}$ `= \frac{2,9375}{6,16875}\approx 0,47` `(l)`