calculate the mass of sodium nitrate needed to make 550 ml of solution of concentration 0;2 mol.dmc to the power of minus 3
To calculate the mass of sodium nitrate needed, we need to use the formula:
mass = concentration × volume × molar mass
1. Firstly, let's calculate the molar mass of sodium nitrate (NaNO3). The atomic masses are as follows: Na (sodium) = 22.99 g/mol, N (nitrogen) = 14.01 g/mol, and O (oxygen) = 16.00 g/mol.
Molar mass of NaNO3 = (22.99 g/mol) + (14.01 g/mol) + 3(16.00 g/mol)
= 22.99 g/mol + 14.01 g/mol + 48.00 g/mol
= 85.00 g/mol
2. Now, substitute the given values into the formula:
mass = (concentration) × (volume) × (molar mass)
mass = (0.2 mol/dm^3) × (550 ml) × (85.00 g/mol)
Note: We need to convert the volume from milliliters to decimeters cubed (dm^3) for consistency in units. There are 1000 mL in 1 dm^3.
Converting volume: 550 ml ÷ 1000 = 0.55 dm^3
mass = (0.2 mol/dm^3) × (0.55 dm^3) × (85.00 g/mol)
3. Calculate the mass:
mass = 9.35 g
Therefore, you would need approximately 9.35 grams of sodium nitrate to make a 550 ml solution with a concentration of 0.2 mol/dm^3.