Calculate the concentration of 5g dextrose in molarity

Molarity of dextrose = 0.03 mol / L

Explanation:

Given data

mass of dextrose = 5 g

Concentration (Molarity) = ?

Solution

as mass is given so we find out number of moles

Moles = mass / molar mass

Molar mass of dextrose = 180.156 g/mol

Moles = 5 g / 180.156 g/mol

Moles = 0.03 moles

Now we find out molarity

Molarity = moles of solute / Litres of solution

As volume is not given so we suppose it as 1 L

Molarity of dextrose = 0.03 mol / 1 L

Molarity of dextrose = 0.03 mol / L

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.