This problem is an acid-base neutralization reaction meaning that the moles of the acid are equivalent to the moles of the base.
The reaction equation for this problem can be written as:
`HCl + NaOH -> NaCl + H_2O`
`Molarity (M) = (mol es of Solute)/(Liters of solution)`
which can also be written as:
`Molarity (M) = (mol es of Solute)/(dm^3 of solution)`
`1 Liter = 1 dm^3`
`M_(NaOH)` = ?
`dm^3` `NaOH =` `20.0 cm^3 (1 dm^3)/(1000 cm^3) = 0.0200 dm^3`
`M_(HCl)` = `1.00 (mol)/(dm^3)`
`dm^3` `HCl =` `10.0 cm^3 (1dm^3)/(1000cm^3) = 0.0100dm^3`
`mol es NaOH = mol es HCl`
`(M_(NaOH))*(dm^3 _N_a_O_H) = (M_(HCl))*(dm^3 _H_C_l)`
`(M_(NaOH))*(0.0200) = (1.0)*(0.0100)`
`M_(NaOH) = ((1.0)*(0.0100))/(0.0200)`
`M_(NaOH) = 0.500 (mol)/(dm^3)`
Th molarity of the NaOH is 0.500mol/dm^3.