a) \(C\%_{dd.sau.khi.trộn}=\dfrac{50.8\%+450.20\%}{50+450}.100\%=18,8\%\)
b) \(V_{dd}=\dfrac{50+450}{1,1}=\dfrac{5000}{11}\left(ml\right)\)
a)
Trong 50 gam dd NaOH có : mNaOH=50.10%=5(gam)
Trong 450 gam dd NaOH có : mNaOH=450.25%=112,5(gam)
Vậy sau khi trộn :
mdd=50+450=500(gam)
mNaOH=5+112,5=117,5(gam)
Suy ra : C%NaOH=117,5\500.100%=23,5%
V{dd}= \(\dfrac{500}{1,1}\)= 454,54(ml)