What is the amount of solute (molar mass = 60 g/mol) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10%?

30 g 60 g 120 g 12 g

60 g

\(\frac{Po - P}{Po}\) = \(\frac{Wb × Ma}{Wa × Mb}\) \(\frac{10}{100}\) = \(\frac{Wb × 18}{180 × 60}\) Wb = \(\frac{10}{100}\) x \(\frac{180 × 60}{18}\) = 60 g