A51.35 g sample of ba(oh)2 (molar mass 171.32 g) is dissolved in enough water to make 1.20 liters of solution. how many ml of this solution must be diluted with water in order to make 1.00 liters of 0.100 m ba(oh)2?

We must diluted 0.40 L in order to make 1 L of 0.1 M Ba(OH)₂

Explanation:

There is something called dilution factor which can help to do this.

Dilution factor is the relationship between diluted and concentrated concentrations, or the quotient  between the volume of the concentrated solution and the volume of the diluted. It also applies to molarity (diluited or concentrated).

Molarity in main solution is: mol/L

Moles = mass / molar mass

51.35 g / 171.32 g/m = 0.299 moles

This moles are in 1.20 L, so in 1 L we have:

0.299 moles/ 1.20 L = 0.249 M

This is the concentrated solution and molarity of diluited is data, 0.1 M.

Dilution factor : diluited/concentrated

0.1 M / 0.249 M = 0.40

Conc. vol. / Diluited vol. = 0.40

Conc. vol / 1 L = 0.40

1 L . 0.40 = Conc. vol = 0.40 L

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.

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