Instruction

1

Mass fraction is the ratio of the mass

Additional formulas which can be necessary for solving problems on the mass fraction

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

2)m = n*M, where m is the mass, n is the number

**of a substance**to the mass of the solution or mixture: w = m ()/m(R-RA), where w – mass fraction, m(in) is the mass of**substance**, m(R-RA) – mass of solution, or w = m ()/m(cm), where m(cm) is the weight of the mixture. Expressed in fraction or percentage.Additional formulas which can be necessary for solving problems on the mass fraction

**of the substance**:1)m = V*p, where m – mass, V – volume, p – density.

2)m = n*M, where m is the mass, n is the number

**of the substance**, M is the molar mass.2

The mole fraction is the ratio of the number of moles

Additional formulas:

1)n = V/Vm, where n – amount

2)n = n/Na where n is the number

**of a substance**to number of moles of all substances: q = n(q)/n(total), where q is the mole fraction, n(q) – the quantity of a specific**substance**, n(total) is the total number of substances.Additional formulas:

1)n = V/Vm, where n – amount

**of substance**V – volume, Vm – molar volume(at normal conditions is equal to 22.4 l/mol).2)n = n/Na where n is the number

**of the substance**, N is the number of molecules, Na – the Avogadro constant(is constant and equal to 6,02*10 23 1/mol).3

The volume fraction is the ratio of the volume

**of the substance**to the volume of the mixture: q = V(q)/V(cm), where q is the volume fraction, V(in) – volume**of the substance**, V(cm) is the volume of the mixture.4

Molar concentration – the ratio of the

Calculated: 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 for a solution volume: V = m/p

m = m(Na2SO4) + m(H2O) = 42,6 + 300 = 342,6 G.

V = 342,6/1,12 = 306 ml = 0,306 L.

Substitute in the General formula: Cm = 0,3/0,306 = 0.98 mol/L. the Problem is solved.

**substance**to the volume of the mixture: Cm = n(V)/V(cm), where Cm is the molar concentration(mol/l), n – amount**of substance**(mol) V(cm) – volume of mixture(l). Solve the problem by the molar**concentration**. Determine the molar**concentration**of a solution obtained by dissolving sodium sulfate with a mass of 42.6 g in water weight of 300 g when the density of the resulting solution is equal to 1.12 g/ml. Write the formula for molar concentration: Cm = n(Na2SO4)/V(cm). See that it is necessary to find the amount of**substance**of sodium and the volume of the solution.Calculated: 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 for a solution volume: V = m/p

m = m(Na2SO4) + m(H2O) = 42,6 + 300 = 342,6 G.

V = 342,6/1,12 = 306 ml = 0,306 L.

Substitute in the General formula: Cm = 0,3/0,306 = 0.98 mol/L. the Problem is solved.

Note

Mass, volume and molar fractions are expressed in fraction or percent, and the molar concentration in mol/L.