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GenaCL600 [577]
2 years ago
9

A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor

r. the vapor pressure of pure water at 30° c is 31.8 torr. calculate the number of moles of nacl
Chemistry
1 answer:
Valentin [98]2 years ago
8 0
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
 So X (the no.of moles of NaCl) = 0.027 m
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lapo4ka [179]

Answer: Concentration of CO = 0.328 M

Concentration of Cl_2 = 0.328 M

Concentration of COCl_2 = 0.532 M

Explanation:

Moles of  CO and Cl_2 = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of CO and Cl_2  =\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M

The given balanced equilibrium reaction is,

                            CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.          0.860M     0.860M           0

At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get :

4.95=\frac{x}{(0.860-x)^2}

By solving the term 'x', we get :

x =  0.532 M

Thus, the concentrations of CO,Cl_2\text{ and }COCl_2 at equilibrium are :

Concentration of CO = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of Cl_2 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M

Concentration of COCl_2 = x M = 0.532 M

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3 years ago
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Answer:

c

Explanation:

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3 years ago
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Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

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4 0
3 years ago
Heat is transferred directly from a heat reservoir at 200 C to another heat reservoir at 5C. If the amount of heat transferred i
Daniel [21]

Answer:

ΔS=0.148  KJ/K

Explanation:

Given that

Q = 100 KJ

T₁=200°C

T₁=200+273 = 437 K

T₂=5°C

T₂=5 + 273 = 278 K

Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.

So the total change in entropy given as

ΔS=  - Q/T₁ + Q/T₂

ΔS=  - 100/473 + 100/278  KJ/K

ΔS=0.148  KJ/K

4 0
3 years ago
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