mNaOH(a%0=100.a/100=a
mNaOH(10%0=50.10/100=5
C%NaOH(7,5)=\(\frac{5+a}{100+50}.100=7,5\)=>a=6,25
K/l NaOh trong dd a%:
m=a/100.100=a(g)
mNaoh trong dd 10%:
50.10/100=5(g)
C% dd naoh 7,5%:
(5+a)/150.100=7,5
Suy ra a=6,25(%)
\(m_{NaOH.10\%}=\frac{50.10}{100}=5\left(g\right)\\ m_{NaOH.a\%}=\frac{100.a}{100}=a\left(g\right)\\ C\%_{ddNaOH.7,5\%}=\frac{a+5}{150}.100=7,5\\ \Leftrightarrow a=\left(\frac{7,5}{100}.150\right)-5=6,25\left(\%\right)\)
mNaOH ( a%) = a (g)
mNaOH ( 10%) = 5 g
mdd = 100 + 50 = 150 g
mNaOH = a + 5 (g)
C%NaOH = (a+5)/150*100% = 7.5%
=> a = 6.25