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

Question

Vapour pressure of CCl4 at 25° C is 143 mm of Hg and 0.5 g of a non-volatile solute (mol. wt. = 65) is dissolved in 100 mL CCl4 . Find the vapour pressure of the solution. (Density of CCl4 =1.58 g text/cm3 ). (A) 94.39 mm (B) 141.93 mm (C) 134.44 mm (D) 199.34 mm

Tardigrade Admin · Accepted Answer

Relative lowering in vapour pressure, (p° -ps/p° )=(w× M/m× W) (143-ps/143)=(0.5/65)× (154/1.58× 100) [ beginalign ∵ Molecular text weight text of CCl4=154 and text weight=density× volume endalign ] 143-ps=1.07 ⇒ ps=141.93 mm

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

94.39 mm

141.93 mm

134.44 mm

199.34 mm

Relative lowering in vapour pressure, p∘p∘−ps​​=m×Ww×M​ 143143−ps​​=650.5​×1.58×100154​ [​∵MolecularweightofCCl4​=154andweight=density×volume​] 143−ps​=1.07 ⇒ ps​=141.93mm

Q. Vapour pressure of $CC l_{4}$ at $25^{\circ} C$ is 143 mm of Hg and 0.5 g of a non-volatile solute (mol. wt. = 65) is dissolved in 100 mL $CC l_{4} .$ Find the vapour pressure of the solution. (Density of $CC l_{4} = 1.58 g / c m^{3}$ ).