5cm^3 of saturated solution of sodium chloride at 30°c give on careful evaporation 1.95g of solid salt, the solubility of sodium chloride at 30°c is
To find the solubility of sodium chloride at 30°C, we first need to convert the mass of the solid salt obtained through evaporation to moles.
The molar mass of sodium chloride (NaCl) is 58.44 g/mol.
1.95 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.0334 mol NaCl
Next, we need to find the volume of water in the saturated solution.
1 cm^3 of water = 1 mL of water
Therefore, 5 cm^3 of saturated solution is equal to 5 mL.
Now, we can calculate the solubility of sodium chloride at 30°C:
Solubility (mol/L) = moles of solute / volume of solvent (L)
Solubility = 0.0334 mol / 0.005 L
Solubility = 6.68 mol/L
Therefore, the solubility of sodium chloride at 30°C is 6.68 mol/L.