Guide

1

Mass fraction - a ratio of weight

additional formulas which may be needed to meet the challenges in the mass fraction of substance

1) m = V * p, where m - mass, V - volume, p - density.

2) m = n * M, where m - mass, n - the number of substances

**substance**to the weight of the solution or mixture: w = m (in) / m (p-pa), where w - mass fraction, m (c) - a substance weight**, m (p-pa) - weight of the solution, or w = m (to) / m (cm), where m (cm) - weight of the blend.It is expressed as a decimal or percentage.**additional formulas which may be needed to meet the challenges in the mass fraction of substance

**:**1) m = V * p, where m - mass, V - volume, p - density.

2) m = n * M, where m - mass, n - the number of substances

**, M - molar mass.** 2

mole fraction - the ratio of moles
tances.Additional

formula:

1) n = V / Vm, where n - the number

2) n = N / Na, where n is the number of substances

**substance**to moles of all substances: q = n (a) / n (tot), where q - mole fraction, n (a) - amount of the particular**substance**, n (total) - the total number of subsformula:

1) n = V / Vm, where n - the number

**substance**, V - volume, Vm - molar volume (under normal conditions is 22.4 liters / mole).2) n = N / Na, where n is the number of substances

**, N - number of molecules, Na - Avogadro constant (a constant and is equal to 6.02 * 10 23 degrees 1 / mol).** 3

volume fraction - is the ratio of

**substance**by volume mixture: q = V (c) / V (cm), where q - volume fraction, V (c) - the volume**substance**,V (cm) - the volume of the mixture. 4

Molar concentration - the ratio of this

expect: n (Na2SO4) = m (Na2SO4) / M (Na2SO4).

M (Na2SO4) = 23 * 2 + 32 + 16 * 4 = 142 g / mol.

n (Na2SO4) = 42,6 / 142 = 0.3 mol.Looking

solution volume: V = m / p

m = m (Na2SO4) + m (H2O) = 42,6 + 300 = 342.6 g = 342.6

V / 1,12 = 306 ml = 0,306 l.

substitute to the general formula: Cm = 0,3 / 0,306 = 0,98 mol / l.The problem is solved.

**substance**by volume mixture: Cm = n (a) / V (cm) where Cm - molar concentration (mol / l), n - the number**substance**(mol), V (cm) - the amount of the mixture (L).We will solve the problem in the molar concentration**.The molar concentration****solution obtained by dissolving sodium sulfate weight of 42.6 g in water of 300 g when the specific gravity of the resulting solution is 1.12 g / ml.Writing a formula to calculate the molar concentration: Cm = n (Na2SO4) / V (cm).We see that the need to find a number of substances****sodium and volume of the solution.**expect: n (Na2SO4) = m (Na2SO4) / M (Na2SO4).

M (Na2SO4) = 23 * 2 + 32 + 16 * 4 = 142 g / mol.

n (Na2SO4) = 42,6 / 142 = 0.3 mol.Looking

solution volume: V = m / p

m = m (Na2SO4) + m (H2O) = 42,6 + 300 = 342.6 g = 342.6

V / 1,12 = 306 ml = 0,306 l.

substitute to the general formula: Cm = 0,3 / 0,306 = 0,98 mol / l.The problem is solved.

Note

mass, volume and mole fractions are expressed as a decimal or percentage, and molar concentration - in mol / l.

Sources:

- "Handbook of Chemistry", GPKhomchenko, 2005.
- formula concentration of the substance