\(n_{HCOOCH_3}=\dfrac{6}{60}=0,1\left(mol\right)\)
PTHH: HCOOCH3 + NaOH --> HCOONa + CH3OH
0,1------->0,1---------->0,1
=> mHCOONa = 0,1.68 = 6,8 (g)
\(n_{NaOH\left(dư\right)}=\dfrac{10,8-6,8}{40}=0,1\left(mol\right)\)
=> nNaOH(bđ) = 0,1 + 0,1 = 0,2 (mol)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,2}{0,1}=2M\)