A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is needed to titrate a 25 mL aliquot of the unknown monoprotic acid. From this info, calculate the molar mass of the unknown acid (g/mol).

36.45 g/mol

Explanation:

The titration of a monoprotic acid with a base as NaOH is:

HX + NaOH → NaX + H₂O.

That means 1 mol of acid reacts per 1 mol of base.

10.0mL of 0.10M NaOH are:

0.0100L ₓ (0.10mol / L) = 0.0010 moles NaOH ≡ 0.0010 moles HX

If 72.9mg of the monoprotic acid are dissolved in 50.0 mL of water, 25 mL of the solution will contains:

72.9mg × (25mL / 50mL) = 36.45mg = 0.03645g of HX.

Molar mass is ratio between mass and moles of substance, that is:

0.03645g of HX / 0.0010 moles HX = 36.45 g/mol

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.