How many grams are needed to make 2 L of solution of 4 M LiNO3?

552 g of LiNO₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mass of LiNO₃ =?

Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mole of LiNO₃ =?

Molarity = mole /Volume

4 = mole of LiNO₃ / 2

Cross multiply

Mole of LiNO₃ = 4 × 2

Mole of LiNO₃ = 8 moles

Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:

Mole of LiNO₃ = 8 moles

Molar mass of LiNO₃ = 7 + 14 + (16×3)

= 7 + 14 + 48

= 69 g/mol

Mass of LiNO₃ =?

Mole = mass /Molar mass

8 = Molar mass of LiNO₃ /69

Cross multiply

Molar mass of LiNO₃ = 8 × 69

Molar mass of LiNO₃ = 552 g

Thus, 552 g of LiNO₃ is needed to prepare the solution.

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

um i think it might be a car?