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Answer :
The vapor pressure of the solution containing 50.0 g of each compound at 296 K is approximately 275.93 torr.
First, we need to calculate the mole fraction of each compound in the solution. The mole fraction is the number of moles of a component divided by the total number of moles of all components in the solution.
The molar mass of carbon disulphide (CS2) is 76.14 g/mol and the molar mass of carbon tetrachloride (CCl4) is 153.82 g/mol.
Number of moles of CS2:
[tex]\[ n_{\text{CS2}} = \frac{50.0 \text{ g}}{76.14 \text{ g/mol}} \approx 0.657 \text{ mol} \][/tex]
Number of moles of CCl4:
[tex]\[ n_{\text{CCl4}} = \frac{50.0 \text{ g}}{153.82 \text{ g/mol}} \approx 0.325 \text{ mol} \][/tex]
Total number of moles in the solution:
[tex]\[ n_{\text{total}} = n_{\text{CS2}} + n_{\text{CCl4}} \approx 0.657 + 0.325 = 0.982 \text{ mol} \][/tex]
Mole fraction of CS2 [tex](CS2):[/tex]
[tex]\[ {\text{CS2}} = \frac{n_{\text{CS2}}}{n_{\text{total}}} \approx \frac{0.657}{0.982} \approx 0.669 \][/tex]
Mole fraction of CCl4 [tex](CCl4):[/tex]
[tex]\[{\text{CCl4}} = \frac{n_{\text{CCl4}}}{n_{\text{total}}} \approx \frac{0.325}{0.982} \approx 0.331 \][/tex]
Now, we can calculate the partial vapor pressures of each component using Raoult's Law:
Partial vapor pressure of CS2 [tex](P_CS2):[/tex]
[tex]\[ P_{\text{CS2}} ={\text{CS2}} \times P_{\text{CS2}}^{\circ} \approx 0.669 \times 360 \text{ torr} \approx 242.84 \text{ torr} \][/tex]
Partial vapor pressure of CCl4 [tex](P_C_C_l_4):[/tex]
[tex]\[ P_{\text{CCl4}} = {\text{CCl4}} \times P_{\text{CCl4}}^{\circ} \approx 0.331 \times 99.8 \text{ torr} \approx 33.09 \text{ torr} \][/tex]
Finally, the total vapor pressure of the solution [tex](P_s_o_l_u_t_i_o_n)[/tex] is the sum of the partial vapor pressures:
[tex]\[ P_{\text{solution}} = P_{\text{CS2}} + P_{\text{CCl4}} \approx 242.84 \text{ torr} + 33.09 \text{ torr} \approx 275.93 \text{ torr} \][/tex]
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Rewritten by : Jeany
Answer:
273.84 Torr
Given
Carbon disulfide = CS2
Carbon tetrachloride = CCL4
CS2 Partial Pressure = 360 Torr
CCL4 Partial Pressure = 99.8 Torr
Grams of CS2 and CCL4 = 50g each
Molar Mass of CS2 = 76.139 g/mol
Molar Mass of CCL4 = 153.82 g/mol
Calculating Moles of both substance...
The molar fraction is calculated by dividing the grams by the molar mass to get moles of each substance
Moles of CS2 = 50g/76.139g/mol
= 0.6567 mol
Moles of CCL4 = 50g/153.82g/mol
= 0.3251 mol
Total = 0.6567 + 0.3251 = 0.9818
Then, we find the Mole fraction of each substance.
Mole Fraction of CS2 = 0.6567/0.9818
= 0.6689
Mole fraction CCL4 = 0.3251/0.9818
= 0.3311
Vapour Pressure of Solution is calculated by
Mole fraction CS2 * Partial pressure CS2 +Mole fraction CCL4 * Partial pressure CCL4
Vapour Pressure = 0.6689 * 360 + 0.3311 * 99.8
= 240.80 + 33.04
= 273.84 Torr
= atm
