Đổi 200ml = 0,2 lít
Ta có: nNaOH = \(\dfrac{16}{40}\) = 0,4 mol
=> CM = \(\dfrac{n}{V}\)= \(\dfrac{0,4}{0,2}\) = 2M
+) Ta có: \(V_{ddNaOH}=200\left(ml\right)=0,2\left(l\right)\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
=> \(C_{MddNaOH}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Ta có:
Vd d NaOH=200 ml =0,2 (lít)
nNaOH=m/M=16/40=0,4(mol)
=> \(C_M=\dfrac{n}{V}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Ta có:
Vd d NaOH=200 ml =0,2 (lít)
nNaOH=m/M=16/40=0,4(mol)
=> CM=nV=0,40,2=2(M)