How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution

0.024 kg CaO

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Aqueous Solutions

Atomic Structure

Step 1: Define

0.41 mol CaO

2.5 M Solution

Step 2: Identify Conversions

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

0.024114 kg CaO ≈ 0.024 kg CaO

0.172 Kg

Explanation:

Given data:

Kg of solvent = ?

Moles of solute = 0.43 mol

Molality = 2.5 m

Solution:

Formula:

m = moles / Kg of solvent

by putting values,

2.5 m = 0.43 mol / kg of solvent

Kg of solvent = 0.43 mol / 2.5 m

Kg of solvent = 0.172 Kg

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.