The solubility of sodium chloride in water is `6.150 mol L^(-1)` at 20°C. 80g of sodium chloride is dissolved in `100 cm^(3)` of water at 20°C. How much sodium chloride is left undissolved ? After equilibrium is reached, an additional 50 cm of water is added to the system at the same temperature. Find the amount of NaCl present in the solution and in the undissolved state.

Explanation

The molecular mass of NaCl = 23 + 35.5
` " " = 58.5`
` " " ` The solubility of NaCl = 6.150 mol/L
` " " = 6.150 xx 58.5`
` " " =359 . 8 g//L`
The mass of NaCl dissolved in ` 100 cm ^(3) ` of water
` " " = (359.8 )/(1000) xx 100`
` " " = 35. 98 g `
Since , 80 g of NaCl is dissolved in ` 100 cm ^(3)` of water the amount of undissolved NaCl = 80 - 35.9 8 = 44.02 g.
Additions of additional ` 50 cm ^(3)` of water will dissolve some more undissolved NaCl . Total volume now becomes ` 150 cm ^(3)`
Hence , the mass of NaCl dissolved in ` 150 cm ^(3)` water
` " " = ( 159. 8)/( 1000) xx 150 = 53. 97 g `
The mass of NaCl still remained undissolved
` " " = 80 - 53. 97 = 26. 03 g `