A solution is prepared by dissolving 10 ~g of a non-volatile solute (molar mass, ' M^ g mol ⁻¹ ) in 360 ~g of water. What is the molar mass in g mol ⁻¹ of solute if the relative lowering of vapour pressure of solution is 5 10⁻³ ?

Step-by-step solution

Given, Mass of solute (w_B )=10 ~g Molar mass of solute (M_B )=M_B Mass of solvent (w_A )=360 ~g Relative lowering in vapour pressure of solution =5 10⁻³ Molar mass of water (M_A )=18 ~g ~mol ⁻¹ p p^ = Relative lower in vapour-pressure of solution. p p^ = _B n_B n_A+n_B =5 10⁻³ where, n_A and n_B are number of moles of solvent (A) and solute (B) respectively. aligned n_A & = 360 18 =20 n_B & = w_B M_B = 10 M_B 5 10⁻³ & = _B= 10 M_B 20+ 10 M_B aligned aligned & 5 10⁻³= 10 M_B 20 M_B+10 M_B & 5 10⁻³= 10 20 M_B+10 & (