The question above is related to these questions along with their answers : Q/ I

Question

The question above is related to these questions along with their answers :

Q/ If 4.798 g of potassium nitrate dissolves in 7.00 g of water at a specific temperature, calculate the solubility of potassium nitrate at this temperature.

A/ m = 4.798 g of KNO3

m = 7 g of water

S = mass/V

assume density f water =) 1g/ml therefore

V = 7 ml = 7*10^-3 L

S = (4.798 )/(7*10^-3 = 685.428 g of KNO3 per liter

Q/ Draw a cartoon diagram (schematic) of the experimental setup. Note on the diagram the volume of water that you will initially use and the approximate mass of the KNO3.

A/ 4.798g/7.00g x 100 = 68.55 KNO3 per 100g of water

Explanation / Answer

The curve has a negative slope. Now, solubility of KNO3 increases with temperature, hence if we had plotted the solubility of KNO3 (in gm/L) vs temeprature (in K), we should have obtained a curve with positive slope. This curve appears to the plot of natural logarithm of solubility product constant (log10Ksp ) against the reciprocal of temperature (1/T) for a solid that undergoes equilibrium while dissolution. Hence, the y-axis represents log10Ksp (no unit) and the x-axis represents 1/T (in K-1). The slope of this plot gives an idea about the enthalpy change (delta H) for the reaction.

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