The density of mercury is 13.5 g/cm3

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

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.

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

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

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

7 0

2 years ago

Someone help me in science plz

NARA [144]

Answer:

I would say Climate - A

Explanation:

Just looks like the logical thing.

7 0

2 years ago

1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.

Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

$W=Fd$

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

$W=(15 N)(0.85 m)=12.75 J$

2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

$W=Fd=12.75 J$