Question

What is the molarity of a solution prepared by dissolving 2.5 grams of LiNO3 in sufficient water to make 60.0 mL of solution?​

Answer 1

Answer:

The molarity of the solution is 0.6 $\frac{moles}{L}$

Explanation:

Molarity (M) is the number of moles of solute that are dissolved in a given volume.

The Molarity is expressed by:

$Molarity (M)=\frac{number of moles of solute}{dissolution volume}$

Molarity is expressed in units $\frac{moles}{liter}$.

Then you must know the moles that represent 2.5 grams of LiNO₃.

For that you must know the molar mass of the compound. You know the atomic mass of the elements that make up the compound, obtained from the periodic table:

• Li: 7 g/mol
• N: 14 g/mol
• O: 16 g/mol

So, taking into account the abundance of each element in the compound:

LiNO₃= 7 g/mol + 14 g/mol + 3*(16 g/mol)= 69 g/mol

Then a rule of three applies as follows: if 69 g of LiNO₃ is 1 mol of the compound, 2.5 g of LiNO₃ how many moles are there?

$moles=\frac{2.5 g*1 mole}{69 g}$

moles= 0.036

Being 60 mL = 0.06 L (1L = 1000 mL or 1 mL = 0.001 L), the molarity is calculated as:

$M=\frac{0.036moles}{0.06L}$

M=0.6 $\frac{moles}{L}$

The molarity of the solution is 0.6 $\frac{moles}{L}$

Answer 2

Answer: hope this helps

To make molar NaCl solutions of other concentrations dilute the mass of salt to 1000ml of solution as follows:

0.1M NaCl solution requires 0.1 x 58.44 g of NaCl = 5.844g.

0.5M NaCl solution requires 0.5 x 58.44 g of NaCl = 29.22g.

2M NaCl solution requires 2.0 x 58.44 g of NaCl = 116.88g.

Explanation: