Question

A laboratory procedure calls for making 410.0mL of a 1.2M NaNO3 solution.

Answer 1

Answer:

The mass of NaNO3 needed is 41.82 grams

Explanation:

The molarity of a compound is defined as the number of moles of the substance in 1 liter of solution. Then, from this definition and using a rule of three, it is possible to calculate the amount of moles present in 410 ml. Remember that 1 L = 1000 ml, then 410 ml = 0.410 L

If 1.2 moles of NaNO3 are present in 1 L of solution, how many moles are in 0.410 L of solution?

$moles of NaNO3=\frac{0.410 L* 1.2 moles}{1 L}$

moles of NaNO3=0.492

And the molar mass of the compound can be calculated by adding the atomic mass of each element present in the compound, taking into account the present amount of each:

• Na: 23 g/mol
• N: 14 g/mol
• O: 16 g/mol

mass of the compound NaNO3= 23 g/mol + 14 g/mol + 16*3 g/mol

mass of the compound NaNO3=85 g/mol

Then, a new rule of three can be applied to calculate the mass of NaNO3 sodium nitrate needed: if in 1 mole they represent 85 grams of sodium nitrate, how many moles are in 0.492 moles?

$necessary mass of NaNO3=\frac{0.492 moles*85 grams}{1 mol}$

necessary mass of NaNO3=41.82 grams

The mass of NaNO3 needed is 41.82 grams

Answer 2

Molar mass NaNO3 = 85 g/mol

Volume in liters: 410.0 mL / 1000 => 0.41 L

mass = M x molar mass x V

mass = 1.2 x 85 x 0.41

mass = 41.82 g of NaNO3