Please show full work and reasoning, thanks! Vapour Pressure of Solutions of Non

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Please show full work and reasoning, thanks!

Vapour Pressure of Solutions of Non-Volatile or Volatile Solutes 9.619 g of a non-volatile solute is dissolved in 310.0 g of water. The solute does not react with water nor dissociate in solution. Assume that the resulting solution displays ideal Raoult's law behaviour. At 25°C the vapour pressure of the solution is 23.451 torr. The vapour pressure of pure water at 25°C is 23.756 torr. Calculate the molar mass of the solute (g/mol) See example 17.1 on pp718-9 of Zumdahl "Chemical Principles" 8th ed. 1pts Submit Answer Tries 0/5 Now suppose, instead, that 9.619 g of a volatile solute is dissolved in 310.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 25°C, a vapour pressure of 2.376 torr. Again, assume an ideal solution. If, at 25°C the vapour pressure of this solution is also 23.451 torr. Calculate the molar mass of this volatile solute. cf p 719 of Zumdahl "Chemical Principles" 8th ed.

Explanation / Answer

According to Raoult’s law:

P = Po*X(solvent)

23.451 = 23.756*X(solvent)

X(solvent) = 0.9872

This is mole fraction of H2O

mass of H2O = 310 g

we have below equation to be used:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(310.0 g)/(18 g/mol)

= 17.22 mol

X(H2O) = n(H2O)/( n(H2O) + n(solute))

0.9872 = 17.22 / ( 17.22+n(solute))

17+0.9872*n(solute) = 17.22

0.9872*n(solute) = 0.2211

n(solute) = 0.224 mol

mass of solute = 9.619 g

we have below equation to be used:

number of mol = mass / molar mass

0.224 mol = (9.619 g)/molar mass

molar mass = 42.94 g/mol

Answer: 42.94 g/mol

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