Question

Calculate the boiling point of a solution containing 0.456 g of camphor (molar mass 152) dissolved in 31.4 gm of acetone (b.pt = 56.30
c.if molal elevation constant per 100 gm of acetone is 17.2 c

Answer 1

Answer:The boiling point of the solution is $56.46^o C$.

Explanation:

Weight of solvent(acetone) = 31.4 g

Molar mass of camphor = 152 g/mol

Mass of camphor added = 0.456 g

$\Delta T_b=100\times K_b\times \frac{\text{mass of camphor}}{\text{molar mass of camphor}\times \text{weight of solvent in g}}$

$K_b=17.2^0C$ per 100 gram of solvent

$\Delta T_b=100\times 17.2\times \frac{0.456 g}{152g/mol\times 31.4g}=0.16^0C$

$\Delta T_b=0.16^0C=T_b-T^{0}_b=T_b-56.30^0C$

$T_b=(56.30+0.16)^0C=56.46^0C$

The boiling point of the solution is $56.46^o C$.