Question

Help please. What is the van’t Hoff factor for a sodium sulfate solution in water that boils at 105oC. The concentration of this solution is 35% by weight sodium sulfate in water. The boiling point elevation constant for water is 0.523oC/m)

Answer 1

use the formula: \(\Delta T=i*m*K_b\)

expand: \((T_f-T_i)=i*\dfrac{n_{solute}}{kg_{solution}}*0.523^oC/m\)

expand further, sub in values : \((105-100)^oC=i*\dfrac{(\dfrac{m_{solute}}{M_{solute}})}{kg_{solution}}*0.523^oC/m\)

assume 100 g of solution, then:\((105-100)^oC=i*\dfrac{(\dfrac{35\;g}{142.04\;g/mol})}{0.065\;kg}*0.523^oC/m\)

solve for i, the van't hoff constant.