A 0.750M solution of H2SO4 in water has a density of 1.046g/mL at 20°C. What is the concentration of this solution in molality g

0.717m

Explanation:

Molality, m, is an unit of concentration defined as the ratio between moles of solute and kg of solvent:

Molality: Moles solute / kg solvent.

In the problem, water is the solvent (Is the compound in the higher quantity) whereas H₂SO₄ is the solute.

The concentration of the solution is 0.750 moles H₂SO₄ / L.

Thus, you need to convert the liters to kg to obtain the molality. That is:

0.750 moles H₂SO₄ / L ₓ (1L / 1000mL) ₓ (1mL / 1.046g) ₓ (1000g / 1kg) =

0.717 moles H₂SO₄ / kg

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.