A 0.623 g sample of a monoprotic acid is dissolved in water and titrated with 0.260 M KOH.
What is the molar mass of the acid if 19.0 mL of the KOH solution is required to neutralize the sample?

126. g/mol is the molar mass of the acid .

Explanation:

Moles of KOH = n

Volume of the solution of KOH = 19.0 mL = 0.019 L

Molarity of the KOH solution = 0.260 M

Accrding to recation, 1 mol KOH react with 1 mole of HA ,then 0.00494 mol of KOH will react with :

of HA

Mass of HA = 0.623 g

Molar mass of HA = M

[tex[M=\frac{0.623 g}{0.00494 mol}=126. g/mol[/tex]

126. g/mol is the molar mass of the acid .

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

what does the table look like