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