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Marysya12 [62]
2 years ago
6

A 52.8-g sample of glucose (a nondissociated, nonvolatile solute with the formula C6H12O6) is dissolved in 158.0 g of water. Wha

t is the vapor pressure of this solution at 100°C?
Chemistry
1 answer:
LUCKY_DIMON [66]2 years ago
7 0
Calculate  the  mole  of  glucose  and  water
 The  moles  of water =158g/18g/mol=8.778 moles
moles of  glucose =52.8g/180g/mol=0.293 moles
determine  the mole  fraction  of  the  solvent
that  is   x solvent = 8.778/ (8.778+0.293)=0.9677
use  the  Raults  law  to  determine  the  vapor  pressure
100  degree  of  water  has   a  vapor  pressure  of  760 mmhg
p solution=(x  solvent) (p^o solvent)
p solution=0.9677 x760 =735.45 mmHg

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1 year ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
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Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

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Answer:

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