Addition of 1.0 g of a compound to 10 g of water .increases the boiling point by 0.3?C. The amount of compound needed to prepare a 500 mL of 0.1 solution is [Given: assume negligible dissociation or association of the compound, boiling point constant kb of water =0.513kg mol-1 ]

Question

Addition of 1.0 g of a compound to 10 g of water .increases the boiling point by 0.3?C. The amount of compound needed to prepare a 500 mL of 0.1 solution is [Given: assume negligible dissociation or association of the compound, boiling point constant kb of water =0.513kg mol-1 ] (A) 0.855 g (B) 17.1 g (C) 8.55 g (D) 85.5 g

Tardigrade Admin · Accepted Answer

Δ Tb=msoluteKb =0.3=(( (1.0/M) )× 1000/10)× 0.513 = 171 [M = molecular weight of solute] We have, Molarity text=(( textWeight/ textM) text ! !× ! ! text 1000/ textVolume textof textsolution(in textmL)) text0 text.1=(( textWeight/171) text ! !× ! ! text 1000/500) Weight = 5.55 g

Q. Addition of 1.0 g of a compound to 10 g of water .increases the boiling point by 0.3?C. The amount of compound needed to prepare a 500 mL of 0.1 solution is [Given: assume negligible dissociation or association of the compound, boiling point constant $k_{b}$ of water $= 0.513 k g m o l^{- 1}$ ]

0.855 g

17.1 g

8.55 g

85.5 g

$Δ T_{b} = m_{so l u t e} K_{b}$ $= 0.3 = \frac{( \frac{1.0}{M} ) \times 1000}{10} \times 0.513$ = 171 [M = molecular weight of solute] We have, Molarity $= \frac{\frac{Weight}{M} \times 1000}{Volume of solution(in mL)}$ $0 .1= \frac{\frac{Weight}{171} \times 1000}{500}$ Weight = 5.55 g

Q. Addition of 1.0 g of a compound to 10 g of water .increases the boiling point by 0.3?C. The amount of compound needed to prepare a 500 mL of 0.1 solution is [Given: assume negligible dissociation or association of the compound, boiling point constant kb​ of water =0.513kgmol−1 ]