1.0 mole NaOH / kg water = 1.0 molal NaOH
The conversions of molality, b, to and from the molarity , c,
for one-solute solutions are:
c = ρ.b / [1 + b.M]
and
b = c / [ρ -c.M]
where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
The molality is 0,2.
0.33 mol/kg (apex)
The conversions of molality, b, to and from the molarity , c,for one-solute solutions are:c = ρ.b / [1 + b.M]andb = c / [ρ -c.M]where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
0.33 mol/kg
0.33 mol/kg
The molality is 0,2.
Molality is moles solute per kilograms of solvent, so 2/10=0.2m.
Molality is defined as moles of solute per kg of solvent. Thus, 3 moles/6 kg = 0.5 moles/kg = 0.5 m
0.33 mol/kg (apex)
The conversions of molality, b, to and from the molarity , c,for one-solute solutions are:c = ρ.b / [1 + b.M]andb = c / [ρ -c.M]where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
0.33 mol/kg
0.33 mol/kg
Molality is expressed as moles solute/kg solvent. Moles of solute = 2. Kg solvent = 6.Molality = 2 moles/6 kg = 0.33 molal
First, you must find the amount of moles of NaOH, using the concentration and volume given. By lowercase m, I'm assuming you mean molality, or molals of solution, which is the equation:molality (m) = (moles of solute) / (total volume of solution (in liters))To solve for moles of NaOH, your solute, rearrange the equation by multiplying volume on both sides to get:moles solute = (molality)(total volume of solution)Next, just plug in the information you know, which is 500 mL for the total volume and 125 m for the molality.***Volume for concentration problems must be converted to liters, so remember that 1 L = 1000 mLmoles NaOH = (125 m)(0.500 L) = 62.5 molesFinally, convert this to grams by finding the molar mass of NaOH using the periodic table:22.99 + 16.00 + 1.008 = 39.998 g/mol62.5 moles (39.998 g) / (1 mol) =249.875 grams NaOH
0.33 mol/kg (apex)
We need to know the Molarity (or Molality or formality) of both the acid and the NaOH solution in order to answer this question.
Molality (m) is defined as moles of solute per kg of solvent.Thus, 2 moles/6 kg = 0.33 molal or 0.3 m.