Question

Determine the boiling point of a solution of 10 grams of a substance of molar mass of 94g/mol in 500 grams of water

Answer 1

Mass of substance$(w)=10g$

Molecular weight of substance$(Gmw)=95g/mol$

Volume of $H_2O(v)=500gms=500ml$

Molarity of solution$(M)=$ $W\over Gmw$ $\times$ $1000\over v(ml)$

$M=$ $10\over 94$ $\times$ $1000\over 500$ $=0.2127M$

Change in boiling point ΔTb​=Kb​×M

$\Delta T_b=0.512°c/_m\times 0.2127M$

$\Delta T_b=0.1089°C$

Boiling point$=100°C+0.1089°C=100.1089°C$