1. what amount of solute present in each of the following solutions?
(a) 750 ml of 2.00 M HNO3 (1.50 moles, 94.5g)
6. Determine Keq for the reaction: 2 SO2 (g) + O2(g) 2 SO3(g), given that 1.00 x 10-2 moles of SO2 and 2.00 x 10-2 moles of O2 were placed in a 2.00L reaction chamber. The chamber contained 7.5 x 10-3 moles of SO3 when equilibrium was established at 727oC.
(PV = nRT, R = 0.0821L• atm/mol • K)
To determine the amount of solute present in the given solution, we need to use the formula:
Amount of solute (g) = concentration (mol/L) × volume (L) × molar mass (g/mol)
Given:
- Volume of solution (V) = 750 mL = 0.75 L (since 1 L = 1000 mL)
- Concentration of HNO3 (C) = 2.00 M (moles per liter)
- Molar mass of HNO3 (M) = 63.01 g/mol (H = 1.01 g/mol, N = 14.01 g/mol, O = 16.00 g/mol)
Using the formula, we can calculate the amount of solute, as follows:
Amount of solute (g) = C × V × M
Plugging in the values:
Amount of solute (g) = 2.00 mol/L × 0.75 L × 63.01 g/mol
Calculating further:
Amount of solute (g) = 94.515 g
Therefore, the amount of solute present in the given solution is approximately 94.5 grams.