Vapour pressure of CCl4 at 25°C is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of CCl4. Find the vapour pressure of the solution. (Density of CCl4 = 1.58 g/cm3)

Question

Vapour pressure of CCl4 at 25°C is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of CCl4. Find the vapour pressure of the solution. (Density of CCl4 = 1.58 g/cm3) (A) 141.93 mm (B) 94.39 mm (C) 199.34 mm (D) 143.99 mm

Tardigrade Admin · Accepted Answer

WB = 0.5 g, MB = 65 amu WA = volume Ã density = 100 Ã 1.58 = 158 g MA(CCl4) = 12 + 35.5 Ã 4 = 154 g mol-1 (p°-ps/po) = (WB/MB)⋅(MA/WA) p°-ps = (WBMA/MBWA) × p° or ps = p° (WBMA/MBWA) × p° = 143-(0.5×154/65×158)×143 = 143 - 1.07 = 141.93 mm

Q. Vapour pressure of CCl4​ at 25∘C is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of CCl4​. Find the vapour pressure of the solution. (Density of CCl4​=1.58g/cm3)

141.93 mm

94.39 mm

199.34 mm

143.99 mm

WB​=0.5g,MB​=65amu WA​= volume × density =100×1.58=158g MA​(CCl4​)=12+35.5×4=154gmol−1 pop∘−ps​​=MB​WB​​⋅WA​MA​​ p∘−ps​=MB​WA​WB​MA​​×p∘ or ps​=p∘MB​WA​WB​MA​​×p∘ =143−65×1580.5×154​×143=143−1.07=141.93mm

Q. Vapour pressure of $CC l_{4}$ at $2 5^{\circ} C$ is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of $CC l_{4}$ . Find the vapour pressure of the solution. (Density of $CC l_{4} = 1.58 g / c m^{3}$ )