A nonvolatile organic compound Z was used to make up a solution. Solution A contains 6.00 g of Zdissolved in 100 g of waterand has a vapor pressure of 754.5 mmHgat the normal boiling point of water. Calculate the molar mass of Z in solution A.

Molar mass of Z in solution A is 148.2 g/mol.

Explanation:

To solve this, we need to use colligative property about vapor pressure.

In any ideal solution, the vapor pressure of solution is lower than vapor pressure of pure solvent. Formula is:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

At the normal boiling point of water (100°C), vapor pressure is 760 mmHg.

Let's replace in the formula, to find out Xm (mole fraction of solute)

760 mmHg - 754.5 mmHg = 760 mmHg . Xm

(760 mmHg - 754.5 mmHg) / 760 mmHg = Xm

Xm → 0.00724

Mole fraction of solute = moles of solute /total moles (st + sv)

In this case, our mole fraction will be

(6 g / MM) / ( (6 g/ MM + 100 g / 18g/mol) = 0.00724

Our unknown is MM (molar mass of Z). We solve the equation:

6 g / MM = 0.00724 . ((6 g/ MM + 100 g / 18g/mol))

6 g / MM =  0.04344 g /MM + 0.0402 g/mol

6 g/MM - 0.04344 g /MM = 0.0402 g/mol

5.95656 /MM = 0.0402 g/mol

5.95656 / 0.0402 g/mol = MM → 148.2 g/mol