Question

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL of H2O at 40.0°C? The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3.

Answer 1

`P_(sol)=50.4\ mm.Hg`

Step-by-step explanation:

According to given:

• molecular mass of glycerin,
  `M_g=3* 12+8+3* 16=92\ g.mol^(-1)`

• molecular mass of water,
  `M_w=2+16=18\ g.mol^(-1)`

• ∵Density of water is
  `0.992\ g.cm^(-3)= 0.992\ g.mL^(-1)`

• ∴mass of water in 316 mL,
  `m_w=316* 0.992=313.5 g`

• mass of glycerin,
  `m_g=154\ g`

• pressure of mixture,
  `P_x=55.32\ torr= 55.32\ mm.Hg`

• temperature of mixture,
  `T_x=40^(\circ)C`

Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.

moles of water in the given quantity:

`n_w=(m_w)/(M_w)`

`n_w=(313.5)/(18)`

`n_w=17.42 moles`

moles of glycerin in the given quantity:

`n_g=(m_g)/(M_g)`

`n_g=(154)/(92)`

`n_g=1.674 moles`

Now the mole fraction of water:

`X_w=(n_w)/(n_w+n_g)`

`X_w=(17.42)/(17.42+1.674)`

`X_w=0.912`

Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.

`\therefore P_(sol)=X_w* P_x`

`\therefore P_(sol)=0.912* 55.32`

`P_(sol)=50.4\ mm.Hg`