How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH 4NO 3 in order to prepare a 0.452 m solution?

Chemistry

1 answer:

Ivahew [28]3 years ago

6 0

Answer:

0.768 kg

Explanation:

Step 1: Given data

Mass of ammonium nitrate (solute): 27.8 g

Molality of the solution (m): 0.452 m

Step 2: Calculate the moles corresponding to 27.8 g of ammonium nitrate

The molar mass of ammonium nitrate is 80.04 g/mol.

$27.8 g \times \frac{1mol}{80.04g} = 0.347mol$

Step 3: Calculate the mass of water

The molality is equal to the moles of solute divided by the kilograms of solvent (water).

$m= \frac{n(solute)}{m(solvent)} \\m(solvent) = \frac{n(solute)}{m} = \frac{0.347mol}{0.452mol/kg} = 0.768 kg$

You might be interested in

What is the mass in grams of 6.022×1023 atoms of mass 16.00 amu?

skelet666 [1.2K]

You need the Avogadro's number. I can't remember exactly the calculation.

6 0

3 years ago

A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta

melomori [17]

Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole

The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited

3 0

3 years ago

Write a balanced net ionic equation for the reaction of aqueous solutions of baking soda (NaHCO3) and acetic acid.(A) HCO3–(aq)

IgorC [24]

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium bicarbonate and acetic acid is given as:

$NaHCO_3(aq.)+CH_3COOH(aq.)\rightarrow CH_3COONa(aq.)+H_2O(l)+CO_2(g)$

Ionic form of the above equation follows:

$Na^+(aq.)+HCO_3^-(aq.)+CH_3COO^-(aq.)+H^+(aq.)\rightarrow CH_3COO^-(aq.)+Na^+(aq.)+H_2O(l)+CO_2(g)$

As, sodium and acetate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$HCO_3^-(aq.)+H^+(aq.)\rightarrow CO_2(g)+H_2O(l)$