50cm\(^3\) of a saturated solution of KNO\(_3\) at 40\(^{\circ}\)C contained 5.05g of the salt. What...
50cm\(^3\) of a saturated solution of KNO\(_3\) at 40\(^{\circ}\)C contained 5.05g of the salt. What is the solubility of KNO\(_3\) at 40\(^{\circ}\)C [K = 39, N = 14, O = 16]
- A)1.0moldm\(^{-3}\)
- B)1.5moldm\(^{-3}\)
- C)2.0moldm\(^{-3}\)
- D)5.0moldm\(^{-3}\)
Correct Answer: A)1.0moldm\(^{-3}\)
Explanation
The solubility of KNO\(_3\) at 40\(^{\circ}\)C can be calculated by finding the concentration of KNO\(_3\) in the saturated solution. To do this, we first need to determine the number of moles of KNO\(_3\) in the 5.05g sample.
To find the number of moles, we can use the formula:
\(n = \frac{m}{M}\)
Where \(n\) is the number of moles, \(m\) is the mass (in grams), and \(M\) is the molar mass.
Using the atomic masses given in the question [K = 39, N = 14, O = 16], we can calculate the molar mass of KNO\(_3\):
\(M(KNO_3) = 39 + 14 + (3 \times 16) = 39 + 14 + 48 = 101g/mol\)
Now we can find the number of moles of KNO\(_3\):
\(n = \frac{5.05g}{101g/mol} = 0.05mol\)
Next, we can find the concentration of KNO\(_3\) in the solution by dividing the number of moles by the volume of the solution. The volume is given as 50cm\(^3\), which is equal to 0.05dm\(^3\) (1dm\(^3\) = 1000cm\(^3\)).
\(Concentration = \frac{0.05mol}{0.05dm^3} = 1.0moldm^{-3}\)
Therefore, the solubility of KNO\(_3\) at 40\(^{\circ}\)C is1.0moldm\(^{-3}\), which corresponds toOption A.