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

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

• molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

• 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, $m_w=316\times 0.992=313.5 g$

• 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:

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

moles of glycerin in the given quantity:

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

Now the mole fraction of water:

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

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

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$