When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by $1℃$ then molecular mass of the solute is ($k_{b}$ for $C_{6}H_{6} = 2.53\ kg - mol^{- 1}$)

Question

Dr. Abubacker · Accepted Answer

(d)
$\mathrm{\Delta}T_{b} = \frac{k_{b} 	imes w 	imes 1000}{m 	imes W}$
$	herefore m = \frac{k_{b} 	imes w 	imes 1000}{\mathrm{\Delta}T_{b} 	imes W} = \frac{2.53 	imes 10 	imes 1000}{1 	imes 100} = 253\ g$

Q. When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by $1℃$ then molecular mass of the solute is ($k_{b}$ for $C_{6}H_{6} = 2.53\ kg - mol^{- 1}$)

Explanation

Reviewed by Dr. Abubacker

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