Find the vapour pressure of solution where 0.5g of a non-volatile solute (Mol.wt =65 g ) is dissolved in 100mLCCl4. Given: Density of CCl4=1.58g/cm3 Vapour pressure of CCl4 at 25° C is 143mmHg .

Question

Find the vapour pressure of solution where 0.5g of a non-volatile solute (Mol.wt =65 g ) is dissolved in 100mLCCl4. Given: Density of CCl4=1.58g/cm3 Vapour pressure of CCl4 at 25° C is 143mmHg . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mass of CCl 4( W )= Volume × density =100 × 1.58=158 g Mass of solute (.w.)=0.5g P0=143mmHg;Ps=? As we know (P0 - PS/P0)=(w × M/m × W) (143 - Ps/143)=(0 . 5 × 154/65 × 158) 143-PS=(0 . 5 × 154 × 143/65 × 158) 143-Ps=1.072 Ps=143-1.072=141.927mmHg

Q. Find the vapour pressure of solution where 0.5g of a non-volatile solute (Mol.wt =65 g ) is dissolved in 100mLCCl4​. Given: Density of CCl4​=1.58g/cm3 Vapour pressure of CCl4​ at 25∘C is 143mmHg .

Mass of CCl4​(W)= Volume × density =100×1.58=158g Mass of solute (w)=0.5g P0​=143mmHg;Ps​=? As we know P0​P0​−PS​​=m×Ww×M​ 143143−Ps​​=65×1580.5×154​ 143−PS​=65×1580.5×154×143​ 143−Ps​=1.072 Ps​=143−1.072=141.927mmHg

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