Chemistry 2002 JAMB Past Questions - Question 16: The solubility of a salt of molar mass 101 g at 20°C is 0.34 mol dm−³. If 3.40g of the salt is dissolved completely in 250cm³ of water in a beaker, the resulting solution is

Correct Answer

C

Explanation

At 20°C, 0.34mol/dm³ of the salt dissolve= 0.34 x 101 = 34.34gm of the salt dissolvedtherefore 34.34gm dissolve is 1,000cm³ if only 250cm³ is the required mass x = 250/1000 x 34.34 = 8.555gmif only 3.40gm of the salt dissolved, the solution is unsaturated. To determine the concentration of the resulting solution after dissolving 3.40 g of the salt in 250 cm³ of water, you can follow these steps:

1. Calculate the number of moles of the salt:
  \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
  \[ \text{Number of moles} = \frac{3.40 \, \text{g}}{101 \, \text{g/mol}} \]

2. Calculate the concentration of the resulting solution:
  \[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
  \[ \text{Concentration} = \frac{\text{Number of moles}}{0.250 \, \text{L}} \]

Now, plug in the values and calculate:

\[ \text{Number of moles} = \frac{3.40 \, \text{g}}{101 \, \text{g/mol}} \approx 0.0337 \, \text{mol} \]

\[ \text{Concentration} = \frac{0.0337 \, \text{mol}}{0.250 \, \text{L}} \approx 0.1348 \, \text{mol/L} \]

So, the resulting solution has a concentration of approximately \(0.1348 \, \text{mol/L}\).