What is the molecular weight of a monoprotic carboxylic acid if 11.20 ml of 0.9635 m sodium hydroxide is required to titrate 1.253 g of this acid? the reaction is represented by following equation. note: in the equation, r represents an unspecified carbon containing structure rco-h (aq)naoh (aq) rco, na (aq) hoh (u) separate experiments suggest the unknown acid is likely pentanoic acid, c4h, co2h. is the unknown pentanoic acid? why or why not?

116.1 a.m.u.

It is not likely that RCOOH is the pentanoic acid

Explanation:

Let's consider the generic neutralization between NaOH and a monoprotic carboxylic acid.

RCOOH(aq) + NaOH(aq) ⇒ RCOONa(aq) + H₂O(l)

The molar ratio of RCOOH to NaOH is 1:1. The moles of RCOOH are:

The molar mass of RCOOH is:

Thus, the molecular weight is 116.1 a.m.u.

Pentanoic acid has the formula C₅H₁₀O₂ with a molecular weight of 102.1 a.m.u. So, it is not likely that RCOOH is the pentanoic acid.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3